<|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 (as in the provided code).

#### 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` (intuitively, changes in direction change distance to center). 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++ implementation (detailed comments)

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

/*
  Geometry + convex hull + tangent-based angular sweep solution.

  Key idea:
  - Reduce each tribe to convex hull.
  - Minimum Java area = minimum circular-cap area among separating lines.
  - Optimal line can be taken supporting a hull vertex.
  - For each vertex p of hull A, sweep valid line directions through p.
  - Compute cap area in O(1) for a candidate direction.
*/

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); }

    // dot and cross
    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 {
        if(x != p.x) return x < p.x;
        return y < p.y;
    }

    coord_t norm2() const { return x * x + y * y; }
    coord_t angle() const { return atan2(y, x); }
    Point perp() const { return Point(-y, x); }
};

// orientation test: +1 left turn, -1 right turn, 0 collinear (eps)
static int ccw(const Point& a, const Point& b, const Point& c) {
    coord_t v = (b - a) ^ (c - a);
    if(-Point::eps <= v && v <= Point::eps) return 0;
    return v > 0 ? 1 : -1;
}

/*
  Convex hull in CCW order, plus split into lower/upper chains for tangent queries.

  The tangent routine below is a known trick:
  - Store hull in CCW order.
  - Maintain "lower" chain (increasing x) and "upper" chain (decreasing x),
    and binary search for places where the turn direction changes from point p.
*/
struct ConvexHull {
    vector<Point> pts;     // CCW hull vertices
    int lower_end = 0;     // index split in pts[]
    vector<Point> lower, upper;

    ConvexHull() {}
    ConvexHull(vector<Point> v) { build(std::move(v)); }

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

    void build(vector<Point> v) {
        sort(v.begin(), v.end());
        v.erase(unique(v.begin(), v.end(), [](const Point& a, const Point& b){
            return a.x == b.x && a.y == b.y;
        }), v.end());

        if((int)v.size() <= 2) { // degenerate hull
            pts = v;
            lower_end = (int)pts.size() - 1;
            lower = pts;
            upper.clear();
            if(!pts.empty()) {
                upper.push_back(pts.back());
                if((int)pts.size() > 1) upper.push_back(pts.front());
            }
            return;
        }

        // Andrew monotone chain but arranged to match the original approach
        vector<int> hull = {0};
        vector<char> used(v.size(), 0);

        auto expand = [&](int i, int min_sz) {
            while((int)hull.size() >= min_sz &&
                  ccw(v[hull[hull.size()-2]], v[hull.back()], v[i]) >= 0) {
                used[hull.back()] = 0;
                hull.pop_back();
            }
            hull.push_back(i);
            used[i] = 1;
        };

        for(int i = 1; i < (int)v.size(); i++) expand(i, 2);
        int uhs = (int)hull.size();

        for(int i = (int)v.size() - 2; i >= 0; i--) {
            if(!used[i]) expand(i, uhs + 1);
        }
        hull.pop_back(); // remove duplicated start

        // build points, then reverse to CCW
        pts.clear();
        for(int id : hull) pts.push_back(v[id]);
        reverse(pts.begin(), pts.end());

        // split into lower/upper chains for tangent binary searches
        lower_end = size() - uhs;
        lower.assign(pts.begin(), pts.begin() + lower_end + 1);
        upper.assign(pts.begin() + lower_end, pts.end());
        upper.push_back(pts[0]); // close chain for convenience
    }

    /*
      Return indices (a,b) of tangent vertices from point p to this hull.
      - a: extreme such that ccw(p, pts[a], pts[id]) is maximized (+ side)
      - b: opposite extreme ( - side)
      Complexity: O(log n) using chain binary searches.
    */
    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, pts[a], pts[id]) > 0) a = id;
            if(ccw(p, pts[b], pts[id]) < 0) b = id;
        };

        auto bin_search = [&](int low, int high) {
            if(low >= high) return;
            update(low);
            int sgn = ccw(p, pts[low % n], pts[(low + 1) % n]);
            while(low + 1 < high) {
                int mid = (low + high) / 2;
                if(ccw(p, pts[mid % n], pts[(mid + 1) % n]) == sgn) 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)
        int uid = (int)(lower_bound(upper.begin(), upper.end(), p,
                    [](const Point& A, const Point& B){
                        // "greater<Point>" lexicographic
                        if(A.x != B.x) return A.x > B.x;
                        return A.y > B.y;
                    }) - upper.begin());

        int base = lower_end;
        bin_search(base, base + uid);
        bin_search(base + uid, base + (int)upper.size() - 1);

        return {a, b};
    }
};

static coord_t R; // radius

// normalize angle to [0, 2pi)
static coord_t norm_ang(coord_t a) {
    const coord_t T = 2 * Point::PI;
    while(a < 0) a += T;
    while(a >= T) a -= T;
    return a;
}

/*
  Area of disk∩half-plane for line through point o with direction alpha.

  Steps:
  - d = (cos alpha, sin alpha)
  - intersect line o + t d with circle |x|=R -> two points q1,q2
  - cap area = 1/2 R^2 * dtheta + 1/2 (q1 x q2),
    where dtheta is CCW angle from q1 to q2.
*/
static coord_t cap_area(const Point& o, coord_t alpha) {
    Point d(cos(alpha), sin(alpha));
    coord_t b = o * d;              // o·d
    coord_t c = o.norm2() - R * R;  // |o|^2 - R^2
    coord_t D = b * b - c;
    if(D < -Point::eps) return 0;   // should not happen for valid inputs
    D = max((coord_t)0, D);
    coord_t sd = sqrt(D);

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

    // CCW angle from q1 to q2: atan2(cross, dot), then normalize
    coord_t dtheta = norm_ang(atan2(q2 ^ q1, q2 * q1));

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

/*
  solve_side(A,B):
  Minimum cap area on the side that contains hull A, among lines that separate A from B.

  Enumerate each vertex p of A as a candidate support point for the separating line.
  For fixed p, sweep all allowed directions alpha for lines through p:
  - forbid directions that would cut inside A at p (based on adjacent edges)
  - forbid directions that would intersect B (based on tangents from p to B)
  Evaluate cap_area at event boundaries and stationary candidates.
*/
static coord_t solve_side(const ConvexHull& A, const ConvexHull& B) {
    coord_t best = numeric_limits<coord_t>::infinity();
    int n = A.size();
    if(n == 0) return 0;

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

    vector<pair<coord_t,int>> events;

    for(int i = 0; i < n; i++) {
        Point p = A[i];
        events.clear();
        events.push_back({0.0, 0}); // sentinel for sweeping

        // stationary candidates: direction perpendicular to p (and opposite)
        if(p.norm2() > Point::eps) {
            coord_t a = p.perp().angle();
            events.push_back({norm_ang(a), 0});
            events.push_back({norm_ang(a + Point::PI), 0});
        }

        int cnt = 0; // number of active forbidden intervals at angle 0

        // ban interval (l,r) on angle circle; use eps to avoid boundary ambiguity
        auto ban = [&](coord_t l, coord_t r) {
            l = norm_ang(l + Point::eps);
            r = norm_ang(r - Point::eps);
            if(l > r) cnt++;        // wrapping interval covers angle 0
            events.push_back({l, +1});
            events.push_back({r, -1});
        };

        // 1) forbid cutting inside A at vertex p
        if(n > 1) {
            // between outgoing edge at i and incoming edge at i (as in editorial)
            ban(edge_ang[i], edge_ang[(i + n - 1) % n] + 2 * Point::PI);
        }

        // 2) forbid directions that would hit B: compute tangents from p to 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(events.begin(), events.end(),
             [](auto &a, auto &b){ return a.first < b.first; });

        // sweep; evaluate at boundary angles where feasibility changes
        for(auto [ang, delta] : events) {
            int was = cnt;
            cnt += delta;
            if(was == 0 || cnt == 0) {
                best = min(best, cap_area(p, ang));
            }
        }
    }

    return best;
}

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

    long long r;
    cin >> r;
    R = (coord_t)r;

    int n, m;
    cin >> n;
    vector<Point> java(n);
    for(int i = 0; i < n; i++) {
        long long x, y;
        cin >> x >> y;
        java[i] = Point((coord_t)x, (coord_t)y);
    }

    cin >> m;
    vector<Point> cpp(m);
    for(int i = 0; i < m; i++) {
        long long x, y;
        cin >> x >> y;
        cpp[i] = Point((coord_t)x, (coord_t)y);
    }

    ConvexHull HJ(java), HS(cpp);

    coord_t minJava = solve_side(HJ, HS);
    coord_t maxJava = Point::PI * R * R - solve_side(HS, HJ);

    cout << fixed << setprecision(15) << minJava << " " << maxJava << "\n";
    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, o): return Point(self.x + o.x, self.y + o.y)
    def __sub__(self, o): return Point(self.x - o.x, self.y - o.y)
    def __mul__(self, k): return Point(self.x * k, self.y * k)

    def dot(self, o):   return self.x * o.x + self.y * o.y
    def cross(self, o): return self.x * o.y - self.y * o.x

    def norm2(self): return self.x * self.x + self.y * self.y
    def angle(self): return math.atan2(self.y, self.x)
    def perp(self):  return Point(-self.y, self.x)

    def __lt__(self, o):
        return (self.x, self.y) < (o.x, o.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 norm_ang(a: float) -> float:
    a %= (2.0 * PI)
    if a < 0:
        a += 2.0 * PI
    return a

class ConvexHull:
    """
    Build convex hull in CCW order.
    Additionally split into lower/upper chains to allow tangent queries from a point.
    """
    def __init__(self, pts):
        pts = sorted(pts)
        # deduplicate
        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.pts = []
        self.lower_end = 0
        self.lower = []
        self.upper = []

        self._build(pts)

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

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

    def _build(self, pts):
        if len(pts) <= 2:
            self.pts = pts[:]
            self.lower_end = len(self.pts) - 1
            self.lower = self.pts[:]
            self.upper = []
            if self.pts:
                self.upper.append(self.pts[-1])
                if len(self.pts) > 1:
                    self.upper.append(self.pts[0])
            return

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

        def expand(i, min_sz):
            while len(hull) >= min_sz and ccw(pts[hull[-2]], pts[hull[-1]], pts[i]) >= 0:
                used[hull[-1]] = False
                hull.pop()
            hull.append(i)
            used[i] = True

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

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

        hull.pop()

        self.pts = [pts[i] for i in hull][::-1]  # CCW
        n = len(self.pts)

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

    def tangents_from(self, p: Point):
        """
        Return indices (a,b) of tangent vertices from point p to this convex hull.
        Mirrors the C++ logic using chain binary searches.
        """
        n = len(self.pts)
        if n <= 1:
            return (0, 0)

        a = 0
        b = 0

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

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

        # lower chain: increasing lexicographic
        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)

        # upper chain: decreasing lexicographic (simulate C++ greater<>)
        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 for line through point o with direction alpha.
    """
    d = Point(math.cos(alpha), math.sin(alpha))

    b = o.dot(d)
    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 + d * (-b - sd)
    q2 = o + d * (-b + sd)

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

def solve_side(R: float, A: ConvexHull, B: ConvexHull) -> float:
    """
    Minimum cap area on A's side among separating lines.
    """
    best = float("inf")
    n = len(A)
    if n == 0:
        return 0.0

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

    for i in range(n):
        p = A[i]
        events = [(0.0, 0)]  # sentinel

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

        cnt = 0

        def ban(l, r):
            nonlocal cnt
            l = norm_ang(l + EPS)
            r = norm_ang(r - EPS)
            if l > r:
                cnt += 1  # wraps around 0
            events.append((l, +1))
            events.append((r, -1))

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

        # forbid intersecting B (use tangents)
        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:
                best = min(best, cap_area(R, p, ang))

    return best

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

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

    HJ = ConvexHull(java)
    HS = ConvexHull(cpp)

    min_java = solve_side(R, HJ, HS)
    max_java = PI * R * R - solve_side(R, HS, HJ)

    sys.stdout.write(f"{min_java:.15f} {max_java:.15f}\n")

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

---

If you want, I can also provide:
- a simpler (but slower) version without `O(log n)` tangents (not suitable for the strict time limit), or
- a diagram-oriented explanation of the forbidden angle intervals and tangent meaning.