## 1) Abridged problem statement

You are given a circle (the island) of radius `r` centered at `(0,0)`. There are `n` points (Java villages) and `m` points (Seeplusplus villages), all strictly inside the circle, and it is guaranteed that **some straight line** can separate all Java points to one side and all Seeplusplus points to the other.

The boundary between the tribes is exactly such a straight line; Java’s land is the intersection of the circle with Java’s half-plane.

Compute:
- the **minimum possible area** of Java’s land,
- the **maximum possible area** of Java’s land,

over all separating lines. Output both areas with sufficient floating precision.

---

## 2) Detailed editorial (how the solution works)

### A. Geometry reformulation

A separating line `L` splits the plane into two half-planes. Java’s land is:

\[
\text{JavaLand}(L) = \text{Disk}(R) \cap H(L)
\]

where `H(L)` is the half-plane containing all Java villages (and containing no Seeplusplus village).

So the area we want for a chosen line is simply the area of a **circular segment/cap** cut from the disk by that line (could be small or large depending on which side we assign to Java).

Thus:
- minimizing Java area = find a separating line that gives the smallest disk∩half-plane area containing Java points.
- maximizing Java area = same but largest.

### B. Convex hull reduction

A half-plane contains a set of points iff it contains their convex hull. Therefore, only the **convex hulls** of the two tribes matter.

Let `A = hull(Java)`, `B = hull(See++)`. Any valid separating line must separate these convex polygons.

### C. Key observation: optimum line passes through a hull vertex

For two convex sets, if you translate a separating line parallel to itself while keeping it separating, the area of the disk-cap on one side changes monotonically with signed distance of the line from the center.

So, for a *fixed direction* (fixed normal), the minimum cap area that still contains `A` and excludes `B` is achieved when the line is “as tight as possible” against one of the polygons, i.e. it becomes **supporting**: it touches a convex hull, hence passes through a vertex (or along an edge, but that can be handled by considering vertices; the minimum will be attained at a vertex due to the strict eps handling).

Therefore, to find the minimum Java area it suffices to:
- pick a vertex `p` of `A`,
- consider all lines through `p` that still separate `A` and `B`,
- take the minimum cap area over those directions,
- then take the minimum over all vertices `p` of `A`.

This is exactly what `solve_side(A,B)` computes.

### D. Parameterizing a line through a fixed point by an angle

Fix a vertex `p` and consider a directed line through `p` with direction unit vector:

\[
\vec d(\alpha) = (\cos\alpha, \sin\alpha)
\]

This line intersects the circle in two points `q1, q2` (if it intersects; it always will here because `p` is inside the circle and the line passes through `p`).

For a given direction, we can compute the area of the disk on the “left” side of the directed line. The solution uses a closed-form “cap area” formula:

1. Solve intersection of the parametric line `p + t d` with circle `|x|=R`:
   - with `b = p·d`, `c = |p|^2 - R^2`, discriminant `D = b^2 - c`.
   - roots: `t = -b ± sqrt(D)`
   - intersection points: `q1 = p + d * (-b - sd)`, `q2 = p + d * (-b + sd)`

2. The cap area for the directed arc from `q1` to `q2` is computed as:
   - sector area `= 0.5 * R^2 * Δθ`
   - triangle area `= 0.5 * (q1 × q2)`
   - with `Δθ` taken as the **CCW angle** from `q1` to `q2` (normalized into `[0,2π)`).

That is what `cap_area(p, alpha)` returns.

### E. Which angles are valid? (forbidden intervals)

Not every line through `p` is allowed: it must separate the two convex hulls and must not cut through `A` itself at vertex `p`.

For each vertex `p` of `A`, we build forbidden direction intervals in angle space and then sweep events.

#### 1) Don’t cut through polygon A at p
At vertex `p = A[i]`, the only directions that keep `A` entirely on one side while passing through `p` are those that do **not** go “inside” the polygon at that vertex.

In a convex polygon, the forbidden wedge is determined by the directions of the two adjacent edges:
- `edge_angle[i] = angle(A[i+1] - A[i])`
- `edge_angle[i-1] = angle(A[i] - A[i-1])` (implemented as `(i+n-1)%n`)

The code bans the interval:
\[
[\; edge\_angle[i],\; edge\_angle[i-1] + 2\pi \;]
\]
(handled carefully with normalization and eps).

#### 2) Don’t cut through polygon B (must separate)
For a line through `p` to have all of `B` strictly on the opposite side, the direction must avoid those that pass “between” the two tangents from point `p` to polygon `B`.

So we need the two tangent points from external point `p` to convex polygon `B`. The code provides:

`B.tangents_from(p) -> (ta, tb)` where:
- one index gives the tangent with maximal ccw orientation,
- the other gives the opposite tangent.

Then the forbidden interval (directions that would intersect/cross B) becomes:
- `angL = angle(B[tb] - p) + π`
- `angR = angle(B[ta] - p)`
and we ban `(angL, angR)` appropriately (with wrap-around).

#### 3) Candidate angles to check
The cap area as a function of `α` is smooth except where the set of valid angles changes (interval endpoints). The minimum over valid angles will occur at:
- boundaries of forbidden intervals, and also potentially at local extrema of cap area.

A local extremum happens when the chord is “balanced” w.r.t. the center, which corresponds to directions perpendicular to vector `p` (because moving the line’s direction changes its distance to origin). The code adds:
- `a = angle(p.perp())` and `a+π`

These two directions are inserted as “events” too.

### F. Sweeping events on the circle of angles

All forbidden intervals are converted into “events” on `[0,2π)`:
- add `(l, +1)`, `(r, -1)` with wrap-around handled by starting `cnt=1` when interval crosses 0.
- also add a dummy `(0,0)` and the candidate extremum angles.

After sorting events by angle, sweep:
- maintain `cnt = number of active bans`
- whenever we are at an event where the state switches to or from `cnt==0`, evaluate `cap_area(p, angle)` and take minimum.

This works because feasible angles form a union of intervals, and minima over a smooth function on an interval occur at endpoints or stationary points; the code injects the stationary candidates as events.

### G. Getting min and max Java area

Let `minA = solve_side(hull(Java), hull(See))` = minimal area Java can have.

For maximum Java area, note:
- A separating line gives Java cap area `S` and See++ cap area `πR² - S`.
- If we swap the roles and compute the minimal area that See++ could get, call it `minB`, then the maximal Java area is `πR² - minB`.

So:
- `maxA = πR² - solve_side(hull(See), hull(Java))`

### H. Complexity

- Building each convex hull: `O((n+m) log(n+m))`.
- For each vertex of `A`, tangent search is `O(log |B|)` (the hull is split into lower/upper chains to binary search).
- Overall: `O(|A| log |B| + |B| log |A|)` plus sorting a constant-sized event list per vertex (still `O(log n)` with small constants).

This is fast enough for strict time limits.

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>
using namespace std;

/*
  Convenience stream operators for pairs and vectors (debug/IO helpers).
*/
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) in >> x;
    return in;
}
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) out << x << ' ';
    return out;
}

/*
  We use double for geometry.
*/
using coord_t = double;

/*
  2D point / vector struct with common operations:
  - arithmetic (+,-,*,/)
  - dot product (*)
  - cross product (^)
  - comparisons for sorting
  - norms/angles/rotations
  - helper geometry: ccw, intersection, circumcenter, etc.
*/
struct Point {
    static constexpr coord_t eps = 1e-9;
    static inline const coord_t PI = acos((coord_t)-1.0);

    coord_t x, y;
    Point(coord_t x = 0, coord_t y = 0) : x(x), y(y) {}

    // Vector arithmetic
    Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
    Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
    Point operator*(coord_t c) const { return Point(x * c, y * c); }
    Point operator/(coord_t c) const { return Point(x / c, y / c); }

    // Dot product
    coord_t operator*(const Point& p) const { return x * p.x + y * p.y; }
    // Cross product (2D scalar cross)
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    // Exact comparisons (used only after sorting integer-input points)
    bool operator==(const Point& p) const { return x == p.x && y == p.y; }
    bool operator!=(const Point& p) const { return x != p.x || y != p.y; }
    bool operator<(const Point& p) const {
        return x != p.x ? x < p.x : y < p.y;
    }
    bool operator>(const Point& p) const {
        return x != p.x ? x > p.x : y > p.y;
    }
    bool operator<=(const Point& p) const {
        return x != p.x ? x < p.x : y <= p.y;
    }
    bool operator>=(const Point& p) const {
        return x != p.x ? x > p.x : y >= p.y;
    }

    // Squared length and length
    coord_t norm2() const { return x * x + y * y; }
    coord_t norm() const { return sqrt(norm2()); }
    // Polar angle of vector
    coord_t angle() const { return atan2(y, x); }

    // Rotate vector by angle a around origin
    Point rotate(coord_t a) const {
        return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
    }

    // Perpendicular vector (rotate +90 degrees)
    Point perp() const { return Point(-y, x); }
    // Unit vector in same direction
    Point unit() const { return *this / norm(); }
    // Unit normal
    Point normal() const { return perp().unit(); }

    // Projection/Reflection (not used in this task)
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    // IO
    friend ostream& operator<<(ostream& os, const Point& p) {
        return os << p.x << ' ' << p.y;
    }
    friend istream& operator>>(istream& is, Point& p) {
        return is >> p.x >> p.y;
    }

    /*
      ccw(a,b,c):
      returns +1 if c is to the left of directed segment a->b,
              -1 if to the right,
               0 if collinear (within eps).
    */
    friend int ccw(const Point& a, const Point& b, const Point& c) {
        coord_t v = (b - a) ^ (c - a);
        if(-eps <= v && v <= eps) return 0;
        else if(v > 0) return 1;
        else return -1;
    }

    // Check if p lies on segment [a,b]
    friend bool point_on_segment(const Point& a, const Point& b, const Point& p) {
        return ccw(a, b, p) == 0 &&
               p.x >= min(a.x, b.x) - eps && p.x <= max(a.x, b.x) + eps &&
               p.y >= min(a.y, b.y) - eps && p.y <= max(a.y, b.y) + eps;
    }

    // Point in triangle (including boundary), by checking consistent orientation
    friend bool point_in_triangle(
        const Point& a, const Point& b, const Point& c, const Point& p
    ) {
        int d1 = ccw(a, b, p);
        int d2 = ccw(b, c, p);
        int d3 = ccw(c, a, p);
        return (d1 >= 0 && d2 >= 0 && d3 >= 0) ||
               (d1 <= 0 && d2 <= 0 && d3 <= 0);
    }

    // Intersection of two non-parallel infinite lines (a1-b1) and (a2-b2)
    friend Point line_line_intersection(
        const Point& a1, const Point& b1, const Point& a2, const Point& b2
    ) {
        return a1 +
               (b1 - a1) * ((a2 - a1) ^ (b2 - a2)) / ((b1 - a1) ^ (b2 - a2));
    }

    // Collinearity of vectors from origin
    friend bool collinear(const Point& a, const Point& b) {
        return abs(a ^ b) < eps;
    }

    // Circumcenter of triangle (unused here)
    friend Point circumcenter(const Point& a, const Point& b, const Point& c) {
        Point mid_ab = (a + b) / 2.0;
        Point mid_ac = (a + c) / 2.0;
        Point perp_ab = (b - a).perp();
        Point perp_ac = (c - a).perp();
        return line_line_intersection(
            mid_ab, mid_ab + perp_ab, mid_ac, mid_ac + perp_ac
        );
    }

    /*
      arc_area: signed area contribution for a circular arc (unused by main solution).
    */
    friend coord_t arc_area(
        const Point& center, coord_t r, const Point& p1, const Point& p2
    ) {
        coord_t theta1 = (p1 - center).angle();
        coord_t theta2 = (p2 - center).angle();
        if(theta2 < theta1 - eps) theta2 += 2 * PI;

        coord_t d_theta = theta2 - theta1;
        coord_t cx = center.x, cy = center.y;
        coord_t area = r * cx * (sin(theta2) - sin(theta1)) -
                       r * cy * (cos(theta2) - cos(theta1)) + r * r * d_theta;
        return area / 2.0;
    }

    // Circle-circle intersection points (unused)
    friend vector<Point> intersect_circles(
        const Point& c1, coord_t r1, const Point& c2, coord_t r2
    ) {
        Point d = c2 - c1;
        coord_t dist = d.norm();

        if(dist > r1 + r2 + eps || dist < abs(r1 - r2) - eps || dist < eps) {
            return {};
        }

        coord_t a = (r1 * r1 - r2 * r2 + dist * dist) / (2 * dist);
        coord_t h_sq = r1 * r1 - a * a;
        if(h_sq < -eps) return {};
        if(h_sq < 0) h_sq = 0;
        coord_t h = sqrt(h_sq);

        Point mid = c1 + d.unit() * a;
        Point perp_dir = d.perp().unit();

        if(h < eps) return {mid};
        return {mid + perp_dir * h, mid - perp_dir * h};
    }

    // Intersection of ray and segment (unused)
    friend optional<Point> intersect_ray_segment(
        const Point& ray_start, const Point& ray_through, const Point& seg_a,
        const Point& seg_b
    ) {
        Point ray_dir = ray_through - ray_start;
        if(ray_dir.norm2() < Point::eps) return {};
        Point seg_dir = seg_b - seg_a;
        coord_t denom = ray_dir ^ seg_dir;
        if(fabs(denom) < eps) return {};
        coord_t t = ((seg_a - ray_start) ^ seg_dir) / denom;
        if(t < eps) return {};
        coord_t s = ((seg_a - ray_start) ^ ray_dir) / denom;
        if(s < eps || s > 1 - eps) return {};
        return ray_start + ray_dir * t;
    }
};

/*
  Simple polygon wrapper, mainly for area; hull inherits from it.
*/
class Polygon {
  public:
    vector<Point> points;

    Polygon() {}
    Polygon(const vector<Point>& points) : points(points) {}

    int size() const { return (int)points.size(); }
    const Point& operator[](int i) const { return points[i]; }
    Point& operator[](int i) { return points[i]; }

    // Signed area via shoelace formula
    coord_t area() const {
        coord_t a = 0;
        for(int i = 0; i < size(); i++) {
            a += points[i] ^ points[(i + 1) % size()];
        }
        return a / 2.0;
    }
};

/*
  Point-in-convex-polygon structure (not used in final solve).
*/
class PointInConvexPolygon {
  private:
    Point min_point;
    vector<Point> points_by_angle;

    // Precompute pivot and sort by polar angle around it
    void prepare() {
        points_by_angle = polygon.points;
        auto min_point_it = min_element(points_by_angle.begin(), points_by_angle.end());
        min_point = *min_point_it;

        points_by_angle.erase(min_point_it);
        sort(points_by_angle.begin(), points_by_angle.end(),
            [&](const Point& a, const Point& b) {
                int d = ccw(min_point, a, b);
                if(d != 0) return d > 0;
                return (a - min_point).norm2() < (b - min_point).norm2();
            }
        );
    }

  public:
    Polygon polygon;
    PointInConvexPolygon(const Polygon& polygon) : polygon(polygon) { prepare(); }

    // O(log n) check by binary searching triangle fan
    bool contains(const Point& p) const {
        int l = 0, r = (int)points_by_angle.size() - 1;
        while(r - l > 1) {
            int m = (l + r) / 2;
            if(ccw(min_point, points_by_angle[m], p) >= 0) l = m;
            else r = m;
        }
        return point_in_triangle(min_point, points_by_angle[l], points_by_angle[r], p);
    }
};

/*
  Convex hull (Andrew monotone chain variant) producing CCW order.
  Additionally splits hull into "lower" and "upper" chains and stores an index
  lower_end to support tangent queries from an external point in O(log n).
*/
class ConvexHull : public Polygon {
  public:
    int lower_end;            // index where lower chain ends in points[]
    vector<Point> lower, upper;

    ConvexHull(const vector<Point>& points) {
        this->points = points;

        // Sort points lexicographically and deduplicate
        sort(this->points.begin(), this->points.end());
        this->points.erase(unique(this->points.begin(), this->points.end()),
                           this->points.end());

        // Small hull cases
        if(this->points.size() <= 2) {
            lower_end = (int)this->points.size() - 1;
            lower = this->points;
            upper = {this->points.back()};
            if(this->points.size() > 1) upper.push_back(this->points.front());
            return;
        }

        // 'hull' stores indices into sorted points; used[] marks indices already in hull
        vector<int> hull = {0};
        vector<bool> used(this->points.size());

        // Helper: maintain convexity by popping while turn is non-right (>=0)
        function<void(int, int)> expand_hull = [&](int i, int min_hull_size) {
            while((int)hull.size() >= min_hull_size &&
                  ccw(this->points[hull[hull.size() - 2]],
                      this->points[hull.back()], this->points[i]) >= 0) {
                used[hull.back()] = false;
                hull.pop_back();
            }
            hull.push_back(i);
            used[i] = true;
        };

        // Build lower chain
        for(int i = 1; i < (int)this->points.size(); i++) expand_hull(i, 2);

        int uhs = (int)hull.size(); // size after lower chain

        // Build upper chain by iterating backwards and adding unused points
        for(int i = (int)this->points.size() - 2; i >= 0; i--) {
            if(!used[i]) expand_hull(i, uhs + 1);
        }

        hull.pop_back(); // last equals first

        // Convert index hull into actual points and reverse to CCW
        vector<Point> pts;
        for(int i: hull) pts.push_back(this->points[i]);
        reverse(pts.begin(), pts.end());
        this->points = std::move(pts);

        // Split into lower and upper chains in this CCW order
        lower_end = size() - uhs;
        lower.assign(this->points.begin(), this->points.begin() + lower_end + 1);
        upper.assign(this->points.begin() + lower_end, this->points.end());
        upper.push_back(this->points[0]); // close the chain for binary searching
    }

    /*
      Find tangent vertex indices (a,b) from external point p to this convex polygon.
      Uses binary searches on lower and upper chains. The returned indices are in
      the polygon's points[] indexing.
    */
    pair<int, int> tangents_from(const Point& p) const {
        int n = size();
        if(n <= 1) return {0, 0};

        int a = 0, b = 0; // candidates for "leftmost" and "rightmost" tangents

        // Update tangent candidates with a new vertex id
        auto update = [&](int id) {
            id %= n;
            if(ccw(p, points[a], points[id]) > 0) a = id;
            if(ccw(p, points[b], points[id]) < 0) b = id;
        };

        // Binary search on a chain segment to find where orientation changes
        auto bin_search = [&](int low, int high) {
            if(low >= high) return;
            update(low);
            int sl = ccw(p, points[low % n], points[(low + 1) % n]);
            while(low + 1 < high) {
                int mid = (low + high) / 2;
                if(ccw(p, points[mid % n], points[(mid + 1) % n]) == sl) low = mid;
                else high = mid;
            }
            update(high);
        };

        // Search on lower chain (increasing x)
        int lid = (int)(lower_bound(lower.begin(), lower.end(), p) - lower.begin());
        bin_search(0, lid);
        bin_search(lid, (int)lower.size() - 1);

        // Search on upper chain (decreasing x), using greater<Point> ordering
        int uid = (int)(lower_bound(upper.begin(), upper.end(), p, greater<Point>()) -
                        upper.begin());
        int base = lower_end;
        bin_search(base, base + uid);
        bin_search(base + uid, base + (int)upper.size() - 1);

        return {a, b};
    }
};

coord_t R; // circle radius

// Normalize angle into [0, 2pi)
coord_t normalize(coord_t x) {
    while(x < 0) x += 2 * Point::PI;
    while(x >= 2 * Point::PI) x -= 2 * Point::PI;
    return x;
}

/*
  Compute area of the circle-cap (disk intersection with half-plane) for a line
  passing through point o with direction alpha (unit vector d=(cos,sin)).
  The returned area corresponds to a consistent directed side (based on CCW angle
  from intersection points).
*/
coord_t cap_area(const Point& o, coord_t alpha) {
    // Direction of the line
    Point dir(cos(alpha), sin(alpha));

    // Solve |o + t*dir|^2 = R^2
    // Since |dir|=1: t^2 + 2*(o·dir)*t + (|o|^2 - R^2)=0
    coord_t b = o * dir;            // o·dir
    coord_t c = o.norm2() - R * R;  // |o|^2 - R^2

    coord_t D = b * b - c;          // discriminant (after dividing by 1)
    if(D < -Point::eps) return 0;   // should not happen for valid inputs
    D = max(D, (coord_t)0);         // clamp small negatives due to precision
    coord_t sd = sqrt(D);

    // Two intersection points with circle
    Point q1 = o + dir * (-b - sd);
    Point q2 = o + dir * (-b + sd);

    // CCW angle from q1 to q2 (using atan2 of cross and dot)
    coord_t dt = normalize(atan2(q2 ^ q1, q2 * q1));

    // Area = sector + triangle = 1/2*R^2*dt + 1/2*(q1 x q2)
    return 0.5 * R * R * dt + 0.5 * (q1 ^ q2);
}

/*
  Compute the minimum possible area of the disk-cap that can contain hull A
  while excluding hull B, assuming the separating line must put A on the chosen side.
*/
coord_t solve_side(const ConvexHull& A, const ConvexHull& B) {
    coord_t ans = numeric_limits<coord_t>::max();
    int n = A.size();

    // Precompute angle of each edge of A
    vector<coord_t> edge_angle(n);
    for(int i = 0; i < n; ++i) {
        edge_angle[i] = (A[(i + 1) % n] - A[i]).angle();
    }

    // Events: (angle, delta). delta modifies "banned counter" cnt.
    vector<pair<coord_t, int>> ev;

    // Enumerate each vertex p of A; optimal supporting line can pass through p
    for(int i = 0; i < n; ++i) {
        Point p = A[i];

        ev.clear();
        ev.emplace_back(0, 0); // ensure there is an event at 0 for wrap handling

        // Add candidate stationary directions: perpendicular to vector p
        if(p.norm2() > Point::eps) {
            coord_t a = p.perp().angle();
            ev.emplace_back(normalize(a), 0);
            ev.emplace_back(normalize(a + Point::PI), 0);
        }

        int cnt = 0; // how many forbidden intervals currently active at angle 0

        // Helper to ban an interval (l,r) on the angle circle, with eps shrinking
        auto ban = [&](coord_t l, coord_t r) {
            l = normalize(l + Point::eps);
            r = normalize(r - Point::eps);

            // If interval wraps around 0, start with cnt=1
            if(l > r) ++cnt;

            // Standard sweep encoding: +1 at l, -1 at r
            ev.emplace_back(l, 1);
            ev.emplace_back(r, -1);
        };

        // Ban directions that would cut inside polygon A at vertex p
        if(n > 1) {
            ban(edge_angle[i], edge_angle[(i + n - 1) % n] + 2 * Point::PI);
        }

        // Ban directions that would not separate from B:
        // find tangents from p to convex hull B, then ban the angular region "towards B"
        auto [ta, tb] = B.tangents_from(p);
        coord_t angL = (B[tb] - p).angle() + Point::PI;
        coord_t angR = (B[ta] - p).angle();
        ban(angL, angR);

        // Sort all event angles
        sort(ev.begin(), ev.end());

        // Sweep and evaluate cap_area at angles where we enter/exit allowed region
        for(auto [ang, d]: ev) {
            int was = cnt;
            cnt += d;

            // If either before or after the event we are allowed (cnt==0),
            // then this event angle is a boundary/candidate worth evaluating.
            if(was == 0 || cnt == 0) {
                ans = min(ans, cap_area(p, ang));
            }
        }
    }
    return ans;
}

int n_ja, n_se;
vector<Point> ja, se;

// Read input and store points as doubles (but read ints for exactness)
void read() {
    int r;
    cin >> r;
    R = r;

    cin >> n_ja;
    ja.resize(n_ja);
    for(auto& p: ja) {
        int x, y;
        cin >> x >> y;
        p.x = x;
        p.y = y;
    }

    cin >> n_se;
    se.resize(n_se);
    for(auto& p: se) {
        int x, y;
        cin >> x >> y;
        p.x = x;
        p.y = y;
    }
}

void solve() {
    /*
      We compute:
      - min area for Java = solve_side(hull(Java), hull(See))
      - max area for Java = total_disk_area - min area for See
                            = pi*R^2 - solve_side(hull(See), hull(Java))
    */
    ConvexHull hja(ja);
    ConvexHull hse(se);

    coord_t min_area = solve_side(hja, hse);
    coord_t max_area = Point::PI * R * R - solve_side(hse, hja);

    cout << fixed << setprecision(15) << min_area << " " << max_area << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1; // single test
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same algorithm, with detailed comments)

```python
import math
from bisect import bisect_left

EPS = 1e-9
PI = math.acos(-1.0)

class Point:
    __slots__ = ("x", "y")
    def __init__(self, x=0.0, y=0.0):
        self.x = float(x)
        self.y = float(y)

    def __add__(self, other): return Point(self.x + other.x, self.y + other.y)
    def __sub__(self, other): return Point(self.x - other.x, self.y - other.y)
    def __mul__(self, k):     return Point(self.x * k, self.y * k)
    def __truediv__(self, k): return Point(self.x / k, self.y / k)

    # dot and cross
    def dot(self, other): return self.x * other.x + self.y * other.y
    def cross(self, other): return self.x * other.y - self.y * other.x

    def norm2(self): return self.x * self.x + self.y * self.y
    def norm(self): return math.hypot(self.x, self.y)
    def angle(self): return math.atan2(self.y, self.x)

    def perp(self): return Point(-self.y, self.x)

    def unit(self):
        n = self.norm()
        return self / n

    def __lt__(self, other):
        # Lexicographic order for hull building
        return (self.x, self.y) < (other.x, other.y)

def ccw(a: Point, b: Point, c: Point) -> int:
    v = (b - a).cross(c - a)
    if -EPS <= v <= EPS:
        return 0
    return 1 if v > 0 else -1

def normalize_angle(x: float) -> float:
    # Map to [0, 2*pi)
    x %= (2.0 * PI)
    if x < 0:
        x += 2.0 * PI
    return x

class ConvexHull:
    """
    Builds a convex hull in CCW order.
    Also stores 'lower' and 'upper' chains + a split index to support tangent search.
    This mirrors the structure used in the C++ code.
    """
    def __init__(self, pts):
        pts = sorted(pts)
        # deduplicate (input has unique points, but still safe)
        uniq = []
        for p in pts:
            if not uniq or (p.x != uniq[-1].x or p.y != uniq[-1].y):
                uniq.append(p)
        pts = uniq
        self.points = pts

        if len(pts) <= 2:
            # trivial hull
            self.lower_end = len(pts) - 1
            self.lower = pts[:]
            self.upper = [pts[-1]] + ([pts[0]] if len(pts) > 1 else [])
            return

        hull = [0]
        used = [False] * len(pts)

        def expand(i, min_size):
            # Maintain convexity with ccw >= 0 pop (same criterion as C++ code)
            while len(hull) >= min_size and ccw(pts[hull[-2]], pts[hull[-1]], pts[i]) >= 0:
                used[hull[-1]] = False
                hull.pop()
            hull.append(i)
            used[i] = True

        # lower
        for i in range(1, len(pts)):
            expand(i, 2)
        uhs = len(hull)

        # upper
        for i in range(len(pts) - 2, -1, -1):
            if not used[i]:
                expand(i, uhs + 1)

        hull.pop()  # remove repeated start

        # Build final CCW point order (reverse like C++)
        final_pts = [pts[i] for i in hull][::-1]
        self.points = final_pts
        n = len(final_pts)

        self.lower_end = n - uhs
        self.lower = final_pts[:self.lower_end + 1]
        self.upper = final_pts[self.lower_end:] + [final_pts[0]]

    def __len__(self):
        return len(self.points)

    def __getitem__(self, i):
        return self.points[i]

    def tangents_from(self, p: Point):
        """
        Return indices (a,b) of tangent points from external point p to this hull.
        Port of the C++ method; relies on binary searching on lower/upper chains.
        """
        n = len(self.points)
        if n <= 1:
            return (0, 0)

        a = 0
        b = 0

        def update(idx):
            nonlocal a, b
            idx %= n
            if ccw(p, self.points[a], self.points[idx]) > 0:
                a = idx
            if ccw(p, self.points[b], self.points[idx]) < 0:
                b = idx

        def bin_search(low, high):
            if low >= high:
                return
            update(low)
            sl = ccw(p, self.points[low % n], self.points[(low + 1) % n])
            while low + 1 < high:
                mid = (low + high) // 2
                if ccw(p, self.points[mid % n], self.points[(mid + 1) % n]) == sl:
                    low = mid
                else:
                    high = mid
            update(high)

        # lower_bound on lower chain by lexicographic order
        lid = bisect_left([(q.x, q.y) for q in self.lower], (p.x, p.y))
        bin_search(0, lid)
        bin_search(lid, len(self.lower) - 1)

        # lower_bound on upper chain by "greater<Point>" order in C++:
        # i.e. decreasing lexicographic. We can bisect on keys (-x,-y).
        upper_keys = [(-q.x, -q.y) for q in self.upper]
        uid = bisect_left(upper_keys, (-p.x, -p.y))
        base = self.lower_end
        bin_search(base, base + uid)
        bin_search(base + uid, base + len(self.upper) - 1)

        return (a, b)

def cap_area(R: float, o: Point, alpha: float) -> float:
    """
    Area of circle cap determined by line through point o with direction alpha.
    Uses quadratic to find chord endpoints q1,q2 and formula:
      area = 1/2 * R^2 * dtheta + 1/2 * (q1 x q2)
    """
    dir = Point(math.cos(alpha), math.sin(alpha))

    b = o.dot(dir)
    c = o.norm2() - R * R
    D = b * b - c
    if D < -EPS:
        return 0.0
    if D < 0.0:
        D = 0.0
    sd = math.sqrt(D)

    q1 = o + dir * (-b - sd)
    q2 = o + dir * (-b + sd)

    dt = normalize_angle(math.atan2(q2.cross(q1), q2.dot(q1)))
    return 0.5 * R * R * dt + 0.5 * q1.cross(q2)

def solve_side(R: float, A: ConvexHull, B: ConvexHull) -> float:
    """
    Minimum possible cap area on A's side among valid separating lines.
    """
    ans = float("inf")
    n = len(A)

    edge_angle = []
    for i in range(n):
        edge = A[(i + 1) % n] - A[i]
        edge_angle.append(edge.angle())

    for i in range(n):
        p = A[i]

        events = []
        events.append((0.0, 0))  # ensure start event

        # stationary candidate directions
        if p.norm2() > EPS:
            a = p.perp().angle()
            events.append((normalize_angle(a), 0))
            events.append((normalize_angle(a + PI), 0))

        cnt = 0  # active ban count at angle 0

        def ban(l, r):
            nonlocal cnt
            l = normalize_angle(l + EPS)
            r = normalize_angle(r - EPS)
            if l > r:
                cnt += 1
            events.append((l, 1))
            events.append((r, -1))

        # forbid cutting A at p
        if n > 1:
            ban(edge_angle[i], edge_angle[(i + n - 1) % n] + 2.0 * PI)

        # forbid directions that would intersect B: use tangents from p to B
        ta, tb = B.tangents_from(p)
        angL = (B[tb] - p).angle() + PI
        angR = (B[ta] - p).angle()
        ban(angL, angR)

        events.sort()

        for ang, d in events:
            was = cnt
            cnt += d
            if was == 0 or cnt == 0:
                ans = min(ans, cap_area(R, p, ang))

    return ans

def main():
    import sys
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    R = float(next(it))

    n = int(next(it))
    ja = [Point(int(next(it)), int(next(it))) for _ in range(n)]

    m = int(next(it))
    se = [Point(int(next(it)), int(next(it))) for _ in range(m)]

    hja = ConvexHull(ja)
    hse = ConvexHull(se)

    min_area = solve_side(R, hja, hse)
    max_area = PI * R * R - solve_side(R, hse, hja)

    sys.stdout.write(f"{min_area:.15f} {max_area:.15f}\n")

if __name__ == "__main__":
    main()
```

---

## 5) Compressed editorial

- Replace each tribe’s villages by its convex hull (`A`, `B`), since half-planes depend only on hulls.
- The area for a separating line is the area of a circular cap (disk ∩ half-plane).
- For a fixed direction, sliding a separating line parallel to itself changes cap area monotonically; thus the optimum occurs when the line is **supporting**, i.e. passes through a hull vertex.
- For each vertex `p` of hull `A`, consider all lines through `p` parameterized by angle `α`.
  - Compute cap area in O(1): intersect line `p + t d(α)` with circle, then `area = ½R²Δθ + ½(q1×q2)`.
  - Determine forbidden angle intervals:
    1) directions that cut inside `A` at `p` (between adjacent edge directions),
    2) directions that intersect `B` (between the two tangents from `p` to `B`).
  - Add also `α` where area can be stationary (directions perpendicular to `p`).
  - Sweep event endpoints on `[0,2π)` with a ban counter; evaluate area at boundary/stationary angles when `ban==0`.
- Minimum over all vertices gives minimal Java area.
- Maximum Java area = `πR² - (minimal area for See++)`, computed by swapping hulls.

