<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

500. Circular Island
Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Andrew has just made a breakthrough in geography: he has found an island that was previously unknown to the civilized world. This island has a shape of perfect circle and it is inhabited by two tribes, Java and Seeplusplus. After a brief contact with the aborigines, Andrew found out that the boundary between the lands of the tribes is a straight line. Moreover, he knows the locations of several Java villages and of several Seeplusplus villages (each is of course located within or on the boundary of the corresponding tribe's land). Now he needs to find out what is the minimal and maximal possible area of Java's land. Help him!
Input
The first line of the input file contains one integer r — the radius of the island (1 ≤ r ≤ 109). The next line contains one integer n () — the number of Java villages. Each of the next n lines contains two integers x and y — the coordinates of Java villages. The next line contains one integer m () — the number of Seeplusplus villages. Each of the next m lines contains two integers x and y — the coordinates of Seeplusplus villages. The center of the island has coordinates (0, 0), each village is within the island and at least r/10 away from the island boundary. No two villages coincide. The input is guaranteed to be valid — there will always be at least one straight line separating the Java villages from the Seeplusplus villages.
Output
Output two floating-point numbers, separated with a space — the minimal and maximal possible area of the Java land. An area will be considered correct if it is within 10-6 relative error of the right answer.
Example(s)
sample input
sample output
6
2
3 4
-3 4
1
0 0
12.389928320447176 56.548667764616276

<|response|>
## 1) Abridged problem statement (concise)

You have a circle (island) of radius `r` centered at `(0,0)`. Inside it are:
- `n` Java villages (points),
- `m` Seeplusplus villages (points),

and it is guaranteed that there exists at least one straight line that separates all Java points to one side and all Seeplusplus points to the other.

The boundary between tribes is exactly such a separating line. Java's land is the intersection of the disk with Java's half-plane.

Compute the **minimum** and **maximum** possible area of Java's land over all valid separating lines.

Output both areas with `1e-6` relative error.

---

## 2) Key observations

1. **Only convex hulls matter.**
   A half-plane contains a set of points iff it contains their convex hull.
   So replace Java points by `HullA`, See++ points by `HullB`.

2. **A separating line defines a circular cap area.**
   Java's land for a line `L` is `Disk(r) ∩ HalfPlane(L)` → a circular segment ("cap"), possibly the larger one.

3. **For a fixed line direction, sliding the line is monotone in area.**
   If you keep the line parallel and slide it while staying separating, the cap area changes monotonically with signed distance from the center.
   Therefore the optimum for that direction happens when the line becomes **supporting** (tight) to a hull.

4. **An optimum line can be chosen to pass through a hull vertex.**
   Supporting lines touch a convex polygon at a vertex (or along an edge). It suffices to consider vertices.

5. **For a fixed vertex `p` of HullA, we can sweep allowed line directions (angles).**
   A line through `p` is parameterized by direction angle `α`. Not all `α` are valid:
   - It must not cut through `HullA` at `p` (only support it).
   - It must place `HullB` strictly on the other side (avoid directions that intersect `HullB`), characterized using **two tangents** from `p` to `HullB`.

6. **Maximum area via complement.**
   If Java gets area `S`, See++ gets `πr² - S`.
   So:
   - `minJava = solve(HullJava, HullSee)`
   - `maxJava = πr² - solve(HullSee, HullJava)`

---

## 3) Full solution approach

### Step A — Build convex hulls
Compute convex hulls `A` (Java) and `B` (See++), in CCW order (Andrew/monotone chain). Complexity: `O((n+m) log(n+m))`.

### Step B — Area of cap for a line through point `p` with direction `α`
Let the line be `p + t * d`, where `d = (cos α, sin α)`.

Find intersections with circle `|x| = r`:

Because `|d|=1`:
\[
|p + td|^2 = r^2 \Rightarrow t^2 + 2(p\cdot d)t + (|p|^2 - r^2)=0
\]
Let:
- `b = p·d`
- `c = |p|² - r²`
- `D = b² - c`

Then:
- `t1 = -b - sqrt(D)`, `t2 = -b + sqrt(D)`
- `q1 = p + t1 d`, `q2 = p + t2 d` (the chord endpoints)

Cap area on a consistent side can be computed as:
\[
\text{area} = \frac{1}{2}r^2\Delta\theta + \frac{1}{2}(q_1 \times q_2)
\]
where `Δθ` is the CCW angle from `q1` to `q2` (normalize to `[0, 2π)`), computed by:
\[
\Delta\theta = \operatorname{atan2}(q_2 \times q_1,\; q_2 \cdot q_1)
\]
then normalized.

This is `O(1)` per evaluation.

### Step C — For each vertex `p` of hull A, find all valid direction angles
We will find forbidden angle intervals and then sweep event endpoints on `[0,2π)`.

#### C1) Forbid cutting inside hull A at vertex p
At vertex `A[i]`, adjacent edge directions are:
- `e1 = angle(A[i+1] - A[i])`
- `e2 = angle(A[i] - A[i-1])` (equivalently edge_angle[i-1])

Directions that go "through the interior" at the vertex are forbidden — this is one angular interval between these edge directions (handled with wrap-around).

#### C2) Forbid intersecting hull B (must separate)
For an external point `p`, the directions for which a line through `p` intersects convex polygon `B` form a continuous forbidden angular interval bounded by the **two tangents** from `p` to `B`.

So:
1. Compute tangent points `(ta, tb)` on `B` from `p`.
2. Convert those tangent rays into a forbidden interval of line directions.

Efficient tangent finding can be done in `O(log |B|)` by binary searching on lower/upper hull chains.

#### C3) Candidate angles to evaluate
The minimum over feasible angle intervals occurs at:
- boundaries of forbidden intervals, and
- stationary points of cap area as a function of direction.

A stationary point occurs when the line direction is perpendicular to vector `p`. So add:
- `α = angle(p.perp())` and `α + π` as candidate event angles.

#### C4) Sweep events
Convert each forbidden interval into two events:
- `(l, +1)` and `(r, -1)`, with wrap-around handled by initializing `cnt` appropriately.
Sort events by angle, sweep maintaining `cnt` = number of active bans.
Whenever we are entering/leaving feasibility (`cnt==0` before or after an event), evaluate `cap_area(p, angle)` and keep the minimum.

The result for all vertices of `A` is `solve_side(A, B)` = minimal possible cap area for side containing `A` while excluding `B`.

### Step D — Produce final answers
- `minJava = solve_side(HullJava, HullSee)`
- `maxJava = πr² - solve_side(HullSee, HullJava)`

---

## 4) C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/geometry/point.hpp>
// #include <coding_library/geometry/polygon.hpp>

using namespace std;

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;
};

using coord_t = double;

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) {}

    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); }

    coord_t operator*(const Point& p) const { return x * p.x + y * p.y; }
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    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;
    }

    coord_t norm2() const { return x * x + y * y; }
    coord_t norm() const { return sqrt(norm2()); }
    coord_t angle() const { return atan2(y, x); }

    Point rotate(coord_t a) const {
        return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
    }

    Point perp() const { return Point(-y, x); }
    Point unit() const { return *this / norm(); }
    Point normal() const { return perp().unit(); }
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    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;
    }

    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;
        }
    }

    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;
    }

    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);
    }

    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));
    }

    friend bool collinear(const Point& a, const Point& b) {
        return abs(a ^ b) < eps;
    }

    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
        );
    }

    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;
    }

    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};
    }

    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;
    }
};

class Polygon {
  public:
    vector<Point> points;

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

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

    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;
    }
};

class PointInConvexPolygon {
  private:
    Point min_point;
    vector<Point> points_by_angle;

    void prepare() {
        points_by_angle = polygon.points;
        vector<Point>::iterator 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();
    }

    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
        );
    }
};

class ConvexHull : public Polygon {
  public:
    int lower_end;
    vector<Point> lower, upper;

    ConvexHull(const vector<Point>& points) {
        this->points = points;
        sort(this->points.begin(), this->points.end());
        this->points.erase(
            unique(this->points.begin(), this->points.end()), this->points.end()
        );

        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;
        }

        vector<int> hull = {0};
        vector<bool> used(this->points.size());

        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;
        };

        for(int i = 1; i < (int)this->points.size(); i++) {
            expand_hull(i, 2);
        }

        int uhs = hull.size();
        for(int i = (int)this->points.size() - 2; i >= 0; i--) {
            if(!used[i]) {
                expand_hull(i, uhs + 1);
            }
        }

        hull.pop_back();

        vector<Point> pts;
        for(int i: hull) {
            pts.push_back(this->points[i]);
        }
        reverse(pts.begin(), pts.end());
        this->points = std::move(pts);

        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]);
    }

    pair<int, int> tangents_from(const Point& p) const {
        int n = size();
        if(n <= 1) {
            return {0, 0};
        }

        int a = 0, b = 0;
        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;
            }
        };

        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);
        };

        int lid =
            (int)(lower_bound(lower.begin(), lower.end(), p) - lower.begin());
        bin_search(0, lid);
        bin_search(lid, (int)lower.size() - 1);

        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;

coord_t normalize(coord_t x) {
    while(x < 0) {
        x += 2 * Point::PI;
    }
    while(x >= 2 * Point::PI) {
        x -= 2 * Point::PI;
    }
    return x;
}

coord_t cap_area(const Point& o, coord_t alpha) {
    Point dir(cos(alpha), sin(alpha));

    coord_t b = o * dir;
    coord_t c = o.norm2() - R * R;

    coord_t D = b * b - c;
    if(D < -Point::eps) {
        return 0;
    }
    D = max(D, (coord_t)0);
    coord_t sd = sqrt(D);

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

    coord_t dt = normalize(atan2(q2 ^ q1, q2 * q1));

    return 0.5 * R * R * dt + 0.5 * (q1 ^ q2);
}

coord_t solve_side(const ConvexHull& A, const ConvexHull& B) {
    coord_t ans = numeric_limits<coord_t>::max();
    int n = A.size();

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

    vector<pair<coord_t, int>> ev;
    for(int i = 0; i < n; ++i) {
        Point p = A[i];

        ev.clear();
        ev.emplace_back(0, 0);

        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;

        auto ban = [&](coord_t l, coord_t r) {
            l = normalize(l + Point::eps);
            r = normalize(r - Point::eps);
            if(l > r) {
                ++cnt;
            }
            ev.emplace_back(l, 1);
            ev.emplace_back(r, -1);
        };

        if(n > 1) {
            ban(edge_angle[i], edge_angle[(i + n - 1) % n] + 2 * Point::PI);
        }

        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(ev.begin(), ev.end());

        for(auto [ang, d]: ev) {
            int was = cnt;
            cnt += d;
            if(was == 0 || cnt == 0) {
                ans = min(ans, cap_area(p, ang));
            }
        }
    }
    return ans;
}

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

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 want to find the min and max area of one tribe's land on a circular
    // island of radius R, separated by a straight line from the other tribe.
    // The key observation is that the optimal separating line must pass through
    // a vertex of one of the two convex hulls.
    //
    // Let's fix a vertex v of hull A and consider all lines through v. Each
    // such line cuts the circle into two caps. We want the minimum cap area on
    // A's side, over all lines that validly separate A from B. As the line
    // rotates around v, the cap area changes, and we can compute it for any
    // given angle: the line hits the circle at two points q1, q2, and the cap
    // area is sector(q1, q2) + triangle(q1, q2) using the directed (CCW) angle.
    //
    // Not all angles are valid though. The line must not cut through either
    // hull. This gives us two forbidden angle intervals:
    //   1) The line can't cut through A itself. Since the hull is CCW, the
    //      forbidden directions are determined by the two edges adjacent to v.
    //   2) The line can't cut through B. We find the two tangent lines from v
    //      to hull B, and the directions between them (on the B side) are
    //      forbidden. Finding these tangent points can be done in O(log n) by
    //      binary searching on the lower and upper parts of the hull.
    //
    // We also add as candidate angles the perpendicular directions from v to
    // the center (where the cap area has local extrema). After collecting all
    // critical angles and sweeping with a ban counter, we evaluate the cap area
    // at every angle where no ban is active.
    //
    // Taking the minimum over all vertices of A gives the minimum area for A's
    // side. By symmetry, the maximum area is pi*R^2 minus the minimum area
    // when we swap the roles of A and B. Total complexity is O((n+m) log(n+m)).

    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;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 5) Python implementation (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()
```
