## 1) Abridged problem statement

A circular pizza of radius **R** centered at (0,0) is cut by **N** straight lines (1 ≤ N ≤ 4), each given by `ax + by + c = 0`. Each line intersects the pizza boundary in exactly two points, so it truly cuts the disk.

Carlsson and Winnie-the-Pooh alternately **take an entire remaining piece** and eat it. They eat at the same speed, and the **time** to eat a piece is **proportional to its area**. Therefore, the next person to choose a new piece is whoever finishes eating earlier; if both finish at the same time and try to take simultaneously, **Carlsson chooses first**.

Assuming both play optimally to maximize their own total eaten area, output the final areas eaten by **Carlsson** and **Pooh**.

Accuracy must be within **1e-4**.

---

## 2) Detailed editorial (geometry + game)

### A. Turn rule as a scheduling game
Because eating time equals area (same speed), each player’s “busy time” is the sum of areas of pieces they already took.

Let:
- `tc` = total area Carlsson has taken so far (thus time spent so far)
- `tp` = total area Pooh has taken so far

At any moment, the next chooser is:
- Carlsson if `tc <= tp` (tie goes to Carlsson),
- otherwise Pooh.

So the game is: from a multiset of piece areas, players pick pieces one by one, but **the next picker is determined by current cumulative sums**, not by strict alternation.

This is not greedy. Taking a huge piece can “stall” you and let the opponent take several smaller pieces before you can pick again.

Since N ≤ 4, the number of pieces is small (≤ 1 + N(N+1)/2 = 11 for lines in a disk arrangement), so we can do exact minimax over subsets.

---

### B. Computing all piece areas: arrangement inside a circle
We need areas of all faces obtained by intersecting:
- the disk boundary circle, and
- N chords (each is the portion of a line inside the disk).

We build a planar embedding of the arrangement via a **DCEL-like half-edge structure**:

**Vertices**
- For each line: its 2 intersection points with the circle are vertices.
- For each pair of lines: if they intersect inside/on the disk, that intersection point is a vertex.

We deduplicate close points (floating error).

**Edges**
1. **Chord pieces on each line:**  
   Collect all vertices lying on that line (its circle endpoints + all internal line-line intersections), sort them along the line, then connect consecutive vertices with straight half-edges.
2. **Circular arcs along the boundary:**  
   Collect all circle-intersection vertices, sort by polar angle around the origin, then connect consecutive ones in CCW order with arc half-edges (and the last to the first).

Each undirected edge becomes two directed **half-edges** with `twin` pointers.

---

### C. Getting faces using half-edge “next” pointers
For each vertex, we sort all outgoing half-edges by their **direction angle** (tangent direction for arcs, segment direction for chords) in CCW order.

For a half-edge `h`, when traversing a face boundary we want to “keep the face on the left”. The standard DCEL rule used here:

`next(h)` = the outgoing half-edge that is immediately **clockwise** from `twin(h)` at the vertex `to(h)`.

Once all `next` pointers are set, walking `h -> next(h) -> next(next(h)) ...` traces one face boundary. Doing this from all unvisited half-edges enumerates every face exactly once.

---

### D. Area of a face boundary (Green’s theorem)
We compute signed area from the boundary, adding contributions per directed half-edge.

For a **straight segment** from `P` to `Q`:
\[
2A += (P_x Q_y - Q_x P_y)
\]
(the standard polygon shoelace increment).

For a **circular arc** centered at origin with radius r, from angle θ1 to θ2 along the chosen direction:
Green’s theorem gives the arc contribution:
\[
2A += r^2 \cdot (\theta_2 - \theta_1)
\]
where the angle difference must follow the arc direction:
- CCW arc: ensure Δθ > 0 by adding 2π if needed,
- CW arc: ensure Δθ < 0 by subtracting 2π if needed.

After traversal:
\[
A = \frac{1}{2}(2A)
\]

The outer (unbounded) face appears with negative signed area; all real pizza pieces are faces **inside the circle**, so we keep only faces with `A > 0`.

This yields a list `areas[]` of piece areas.

---

### E. Optimal play via minimax with alpha-beta
Let `p = len(areas)` (p ≤ ~11).

State:
- `mask`: which pieces are already taken
- `tc`, `tp`: cumulative eaten areas (times)
- `csf`: Carlsson’s total area so far

Transition:
- Determine whose turn: `tc <= tp` => Carlsson, else Pooh.
- Current player chooses any remaining piece `i`:
  - If Carlsson: new `(tc+ai, tp, csf+ai)`
  - If Pooh: new `(tc, tp+ai, csf)`

Terminal:
- all taken (`mask == (1<<p)-1`) return `csf`.

This is a zero-sum game in terms of Carlsson’s final area: Pooh minimizes it. So:
- Carlsson nodes: take **max**
- Pooh nodes: take **min**

With p small, brute force is fine; alpha-beta pruning helps.

Finally:
- `carlsson = minimax(...)`
- `pooh = total_area - carlsson`

---

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

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

// Pretty-print pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read pair from input
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read a whole vector by reading each element
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) in >> x;
    return in;
}

// Print vector elements separated by spaces
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) out << x << ' ';
    return out;
}

using coord_t = double;

// 2D point/vector with common geometry operations
struct Point {
    static constexpr coord_t eps = 1e-9;              // epsilon for comparisons
    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)
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    // Lexicographic comparisons (mostly unused here)
    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 in [-pi, pi]
    coord_t angle() const { return atan2(y, x); }

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

    // Perpendicular vector (CCW)
    Point perp() const { return Point(-y, x); }

    // Unit vector
    Point unit() const { return *this / norm(); }

    // Unit normal (perpendicular and normalized)
    Point normal() const { return perp().unit(); }

    // Project p onto this vector (interpreting this as direction)
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }

    // Reflect p across this vector line through origin
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    // Stream operators for Point
    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 test: +1 left turn, -1 right turn, 0 collinear
    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 ab (with eps)
    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;
    }

    // Check if p is inside or on triangle abc
    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 point of two non-parallel infinite lines a1b1 and a2b2
    friend Point line_line_intersection(
        const Point& a1, const Point& b1, const Point& a2, const Point& b2
    ) {
        // Uses cross products; assumes denominator != 0
        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;
    }

    // Circumcenter of triangle abc (unused in solution)
    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);
    }

    // Area under a circular arc (unused in final; kept from library)
    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 (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};
    }

    // Ray-segment intersection (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;
    }
};

void read() {}

// Main solver
void solve() {
    // Build arrangement inside a circle using a half-edge structure, compute
    // all bounded faces (pizza slices) areas, then run minimax (alpha-beta)
    // with "who moves next" determined by current cumulative times (areas).

    int n;
    double r;
    cin >> n >> r;

    // Each line stored as (a,b,c) for ax+by+c=0
    vector<array<double, 3>> lines(n);
    for(auto& l: lines) cin >> l[0] >> l[1] >> l[2];

    // Compute the two intersection points between line i and circle radius r
    auto line_circle_pts = [&](int i) -> array<Point, 2> {
        double a = lines[i][0], b = lines[i][1], c = lines[i][2];
        double n2 = a * a + b * b;              // |(a,b)|^2
        // Foot of perpendicular from origin to the line (closest point to origin)
        Point foot(-a * c / n2, -b * c / n2);
        // Distance^2 from foot to intersection points along the line direction
        double h2 = r * r - c * c / n2;
        double h = sqrt(max(0.0, h2));          // guard numerical negatives
        // Unit direction along the line (perpendicular to normal (a,b))
        Point dir = Point(-b, a) / sqrt(n2);
        return {foot + dir * h, foot - dir * h};
    };

    // Compute intersection point of lines i and j, but only if inside the disk
    auto line_line_pt = [&](int i, int j) -> optional<Point> {
        double a1 = lines[i][0], b1 = lines[i][1], c1 = lines[i][2];
        double a2 = lines[j][0], b2 = lines[j][1], c2 = lines[j][2];
        double d = a1 * b2 - a2 * b1;           // determinant
        if(fabs(d) < Point::eps) return nullopt; // parallel (or nearly)
        // Solve linear system
        Point p((-c1 * b2 + b1 * c2) / d, (-a1 * c2 + a2 * c1) / d);
        // Keep only intersections within the pizza disk
        if(p.norm() > r + Point::eps) return nullopt;
        return p;
    };

    // Global vertex list
    vector<Point> verts;

    // Add vertex with dedup by distance; return its index
    auto add_vertex = [&](Point p) {
        for(int i = 0; i < (int)verts.size(); i++) {
            if((verts[i] - p).norm() < 1e-7) return i; // same point
        }
        verts.push_back(p);
        return (int)verts.size() - 1;
    };

    // Half-edge record
    struct HalfEdge {
        int from, to;       // directed edge from -> to
        bool is_arc;        // false: straight segment, true: circle arc
        bool ccw;           // for arcs: direction along circle (ccw or cw)
        int twin;           // index of opposite half-edge
        int next;           // next half-edge around the face
    };

    vector<HalfEdge> hes;

    // Add an undirected edge as two half-edges
    auto add_edge = [&](int u, int v, bool is_arc, bool ccw_uv) {
        int h1 = (int)hes.size();
        // forward half-edge
        hes.push_back({u, v, is_arc, ccw_uv, h1 + 1, -1});
        // backward half-edge is twin; reverse direction, reverse arc direction
        hes.push_back({v, u, is_arc, !ccw_uv, h1, -1});
    };

    // 1) Add all chord segments: for each line, connect consecutive vertices on it
    for(int i = 0; i < n; i++) {
        vector<int> on_line;

        // Add the two circle intersection endpoints
        auto cp = line_circle_pts(i);
        on_line.push_back(add_vertex(cp[0]));
        on_line.push_back(add_vertex(cp[1]));

        // Add all interior intersections with other lines
        for(int j = 0; j < n; j++) {
            if(j == i) continue;
            auto pt = line_line_pt(i, j);
            if(pt) on_line.push_back(add_vertex(*pt));
        }

        // Sort points along the line using projection onto a direction vector
        double a = lines[i][0], b = lines[i][1];
        sort(on_line.begin(), on_line.end(), [&](int u, int v) {
            // direction along line is (-b,a); compare dot with that direction
            return -b * verts[u].x + a * verts[u].y <
                   -b * verts[v].x + a * verts[v].y;
        });

        // Remove duplicates (same vertex index)
        on_line.erase(unique(on_line.begin(), on_line.end()), on_line.end());

        // Connect consecutive points by straight edges (segments)
        for(int k = 0; k + 1 < (int)on_line.size(); k++) {
            add_edge(on_line[k], on_line[k + 1], false, false);
        }
    }

    // 2) Add circle boundary arcs between consecutive circle-intersection points
    vector<int> circle_verts;
    for(int i = 0; i < n; i++) {
        auto cp = line_circle_pts(i);
        circle_verts.push_back(add_vertex(cp[0]));
        circle_verts.push_back(add_vertex(cp[1]));
    }

    // Helper: angle in [0,2pi)
    auto angle_of = [&](int v) {
        double a = atan2(verts[v].y, verts[v].x);
        if(a < 0) a += 2 * Point::PI;
        return a;
    };

    // Sort boundary vertices by polar angle and dedup
    sort(circle_verts.begin(), circle_verts.end(), [&](int u, int v) {
        return angle_of(u) < angle_of(v);
    });
    circle_verts.erase(unique(circle_verts.begin(), circle_verts.end()), circle_verts.end());

    // Connect around the circle in CCW direction
    int cm = (int)circle_verts.size();
    for(int k = 0; k < cm; k++) {
        add_edge(circle_verts[k], circle_verts[(k + 1) % cm], true, true);
    }

    // For sorting outgoing half-edges at a vertex, we need a direction vector
    auto outgoing_dir = [&](int h) -> Point {
        Point p = verts[hes[h].from];
        if(!hes[h].is_arc) {
            // straight segment direction is just to-from
            return verts[hes[h].to] - p;
        }
        // for arcs on circle centered at origin:
        // tangent direction at point p is perpendicular to radius vector (x,y)
        return hes[h].ccw ? Point(-p.y, p.x)   // CCW tangent
                          : Point(p.y, -p.x);  // CW tangent
    };

    // Build adjacency list: outgoing half-edges from each vertex
    int vtotal = (int)verts.size();
    vector<vector<int>> out_at(vtotal);
    for(int h = 0; h < (int)hes.size(); h++) {
        out_at[hes[h].from].push_back(h);
    }

    // Sort outgoing edges CCW by their direction angle
    for(auto& lst: out_at) {
        sort(lst.begin(), lst.end(), [&](int a, int b) {
            Point da = outgoing_dir(a);
            Point db = outgoing_dir(b);
            return atan2(da.y, da.x) < atan2(db.y, db.x);
        });
    }

    // pos_of[h] = index of h inside out_at[from(h)] after sorting
    vector<int> pos_of(hes.size());
    for(int v = 0; v < vtotal; v++) {
        for(int k = 0; k < (int)out_at[v].size(); k++) {
            pos_of[out_at[v][k]] = k;
        }
    }

    // Set next pointers using DCEL rule:
    // next(h) = edge leaving vertex to(h) that is immediately clockwise from twin(h).
    for(int h = 0; h < (int)hes.size(); h++) {
        int t = hes[h].twin;         // go to twin
        int v = hes[t].from;         // vertex we arrive at (same as to(h))
        int k = (int)out_at[v].size();
        // clockwise means previous in CCW-sorted list => index - 1
        hes[h].next = out_at[v][(pos_of[t] - 1 + k) % k];
    }

    // Traverse all faces; each face corresponds to a cycle of next pointers
    vector<bool> visited(hes.size(), false);
    vector<double> areas;

    for(int h0 = 0; h0 < (int)hes.size(); h0++) {
        if(visited[h0]) continue;

        double area2 = 0;  // we accumulate 2*area (signed)
        int cur = h0;

        while(!visited[cur]) {
            visited[cur] = true;

            Point pf = verts[hes[cur].from];
            Point pt = verts[hes[cur].to];

            if(!hes[cur].is_arc) {
                // shoelace increment for segment
                area2 += pf.x * pt.y - pt.x * pf.y;
            } else {
                // arc contribution: r^2 * dtheta (signed according to direction)
                double th1 = atan2(pf.y, pf.x);
                double th2 = atan2(pt.y, pt.x);
                double dth = th2 - th1;

                if(hes[cur].ccw) {
                    // want positive dth in (0, 2pi]
                    if(dth <= 1e-12) dth += 2 * Point::PI;
                } else {
                    // want negative dth in [-2pi, 0)
                    if(dth >= -1e-12) dth -= 2 * Point::PI;
                }
                area2 += r * r * dth;
            }

            cur = hes[cur].next; // follow boundary
        }

        double a = area2 / 2.0;     // signed area
        if(a > 1e-9) {              // keep only bounded positive-area faces (pieces)
            areas.push_back(a);
        }
    }

    // Number of pieces
    int p = (int)areas.size();

    // Minimax with alpha-beta pruning.
    // State: mask of taken pieces, tc/tp cumulative times, csf Carlsson-so-far
    function<double(int, double, double, double, double, double)> best_carlsson =
        [&](int mask, double tc, double tp, double csf, double alpha, double beta) -> double {

        // Terminal: all pieces taken
        if(mask == (1 << p) - 1) return csf;

        // Whose turn? tie goes to Carlsson
        bool c_turn = (tc <= tp + Point::eps);

        if(c_turn) {
            // Carlsson maximizes his final total
            double best = -1e18;
            for(int i = 0; i < p; i++) {
                if(mask & (1 << i)) continue;

                double v = best_carlsson(
                    mask | (1 << i),
                    tc + areas[i],   // Carlsson spends time equal to area
                    tp,
                    csf + areas[i],  // Carlsson gains area
                    alpha, beta
                );

                best = max(best, v);
                alpha = max(alpha, best);
                if(alpha >= beta) break; // prune
            }
            return best;
        } else {
            // Pooh minimizes Carlsson's final total (zero-sum)
            double best = 1e18;
            for(int i = 0; i < p; i++) {
                if(mask & (1 << i)) continue;

                double v = best_carlsson(
                    mask | (1 << i),
                    tc,
                    tp + areas[i],   // Pooh spends time
                    csf,             // Carlsson gains nothing on Pooh's move
                    alpha, beta
                );

                best = min(best, v);
                beta = min(beta, best);
                if(alpha >= beta) break; // prune
            }
            return best;
        }
    };

    // Total pizza area is sum of all piece areas (should equal pi*r^2)
    double total = 0;
    for(double a: areas) total += a;

    // Compute optimal Carlsson area, Pooh gets the rest
    double carlsson = best_carlsson(0, 0, 0, 0, -1e18, 1e18);
    double pooh = total - carlsson;

    cout << fixed << setprecision(10) << carlsson << ' ' << pooh << '\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 approach, detailed comments)

```python
import math
from typing import List, Tuple, Optional

EPS = 1e-9
PI = math.pi

class Point:
    __slots__ = ("x", "y")
    def __init__(self, x: float = 0.0, y: float = 0.0):
        self.x = x
        self.y = 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, c: float): return Point(self.x * c, self.y * c)
    def __truediv__(self, c: float): return Point(self.x / c, self.y / c)

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

    def norm2(self) -> float: return self.x * self.x + self.y * self.y
    def norm(self) -> float: return math.hypot(self.x, self.y)

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

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

def atan2pi(y: float, x: float) -> float:
    """Angle in [0, 2pi)."""
    a = math.atan2(y, x)
    if a < 0:
        a += 2 * PI
    return a

class HalfEdge:
    __slots__ = ("frm", "to", "is_arc", "ccw", "twin", "nxt")
    def __init__(self, frm: int, to: int, is_arc: bool, ccw: bool, twin: int):
        self.frm = frm
        self.to = to
        self.is_arc = is_arc
        self.ccw = ccw
        self.twin = twin
        self.nxt = -1

def solve() -> None:
    import sys
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))
    r = float(next(it))

    # Read lines as (a,b,c)
    lines: List[Tuple[float, float, float]] = []
    for _ in range(n):
        a = float(next(it)); b = float(next(it)); c = float(next(it))
        lines.append((a, b, c))

    # Compute line-circle intersection points for line i
    def line_circle_pts(i: int) -> Tuple[Point, Point]:
        a, b, c = lines[i]
        n2 = a*a + b*b
        # closest point from origin to line
        foot = Point(-a*c/n2, -b*c/n2)
        # distance^2 along line direction to circle intersections
        h2 = r*r - (c*c)/n2
        if h2 < 0:  # numerical guard; problem guarantees intersection
            h2 = 0.0
        h = math.sqrt(h2)
        # direction along line
        dirv = Point(-b, a) / math.sqrt(n2)
        return foot + dirv*h, foot - dirv*h

    # Compute intersection of lines i and j if not parallel and within disk
    def line_line_pt(i: int, j: int) -> Optional[Point]:
        a1, b1, c1 = lines[i]
        a2, b2, c2 = lines[j]
        d = a1*b2 - a2*b1
        if abs(d) < EPS:
            return None
        x = (-c1*b2 + b1*c2) / d
        y = (-a1*c2 + a2*c1) / d
        p = Point(x, y)
        if p.norm() > r + EPS:
            return None
        return p

    # Vertex list with deduplication
    verts: List[Point] = []

    def add_vertex(p: Point) -> int:
        for idx, q in enumerate(verts):
            if (q - p).norm() < 1e-7:
                return idx
        verts.append(p)
        return len(verts) - 1

    hes: List[HalfEdge] = []

    def add_edge(u: int, v: int, is_arc: bool, ccw_uv: bool) -> None:
        h1 = len(hes)
        # forward
        hes.append(HalfEdge(u, v, is_arc, ccw_uv, h1 + 1))
        # backward (twin)
        hes.append(HalfEdge(v, u, is_arc, not ccw_uv, h1))

    # Add chord segments for each line
    for i in range(n):
        on_line: List[int] = []
        p1, p2 = line_circle_pts(i)
        on_line.append(add_vertex(p1))
        on_line.append(add_vertex(p2))
        for j in range(n):
            if j == i:
                continue
            p = line_line_pt(i, j)
            if p is not None:
                on_line.append(add_vertex(p))

        a, b, _ = lines[i]

        # sort along direction (-b, a) using dot product
        on_line.sort(key=lambda vid: (-b)*verts[vid].x + a*verts[vid].y)

        # unique vertex ids (sorted list)
        uniq = []
        last = None
        for vid in on_line:
            if last is None or vid != last:
                uniq.append(vid)
            last = vid
        on_line = uniq

        # connect consecutive vertices
        for k in range(len(on_line) - 1):
            add_edge(on_line[k], on_line[k+1], is_arc=False, ccw_uv=False)

    # Add boundary arcs between consecutive circle intersection vertices
    circle_verts: List[int] = []
    for i in range(n):
        p1, p2 = line_circle_pts(i)
        circle_verts.append(add_vertex(p1))
        circle_verts.append(add_vertex(p2))

    circle_verts.sort(key=lambda vid: atan2pi(verts[vid].y, verts[vid].x))
    uniq = []
    last = None
    for vid in circle_verts:
        if last is None or vid != last:
            uniq.append(vid)
        last = vid
    circle_verts = uniq

    m = len(circle_verts)
    for k in range(m):
        add_edge(circle_verts[k], circle_verts[(k+1) % m], is_arc=True, ccw_uv=True)

    # Direction used for ordering outgoing half-edges at a vertex
    def outgoing_dir(h: int) -> Point:
        p = verts[hes[h].frm]
        if not hes[h].is_arc:
            return verts[hes[h].to] - p
        # tangent to circle at p; direction depends on arc direction
        if hes[h].ccw:
            return Point(-p.y, p.x)
        else:
            return Point(p.y, -p.x)

    # Build outgoing lists per vertex
    out_at: List[List[int]] = [[] for _ in range(len(verts))]
    for h in range(len(hes)):
        out_at[hes[h].frm].append(h)

    # Sort outgoing edges by polar angle of their direction vector
    for v in range(len(verts)):
        out_at[v].sort(key=lambda h: math.atan2(outgoing_dir(h).y, outgoing_dir(h).x))

    # Position lookup for each half-edge in its vertex outgoing list
    pos_of = [0] * len(hes)
    for v in range(len(verts)):
        for k, h in enumerate(out_at[v]):
            pos_of[h] = k

    # Set next pointers: next(h) = edge clockwise from twin(h) at the destination
    for h in range(len(hes)):
        t = hes[h].twin
        v = hes[t].frm
        k = len(out_at[v])
        hes[h].nxt = out_at[v][(pos_of[t] - 1) % k]

    # Traverse faces and compute positive areas
    visited = [False] * len(hes)
    areas: List[float] = []

    for h0 in range(len(hes)):
        if visited[h0]:
            continue
        area2 = 0.0
        cur = h0
        while not visited[cur]:
            visited[cur] = True
            pf = verts[hes[cur].frm]
            pt = verts[hes[cur].to]

            if not hes[cur].is_arc:
                # segment shoelace contribution
                area2 += pf.x * pt.y - pt.x * pf.y
            else:
                # arc contribution: r^2 * dtheta with sign
                th1 = math.atan2(pf.y, pf.x)
                th2 = math.atan2(pt.y, pt.x)
                dth = th2 - th1
                if hes[cur].ccw:
                    if dth <= 1e-12:
                        dth += 2 * PI
                else:
                    if dth >= -1e-12:
                        dth -= 2 * PI
                area2 += r * r * dth

            cur = hes[cur].nxt

        a = area2 / 2.0
        if a > 1e-9:
            areas.append(a)

    p = len(areas)

    # Alpha-beta minimax returning Carlsson's final area
    sys.setrecursionlimit(10000)

    from functools import lru_cache

    @lru_cache(None)
    def dfs(mask: int, tc_q: int, tp_q: int) -> float:
        """
        Memoized version with quantized tc/tp is hard; instead we do
        non-memo alpha-beta in C++. Here we keep it simple without memo:
        but we still need tc/tp as floats; caching floats is unreliable.
        So we implement a direct alpha-beta recursion without caching below.
        """
        return 0.0  # unused

    def best_carlsson(mask: int, tc: float, tp: float, csf: float,
                      alpha: float, beta: float) -> float:
        if mask == (1 << p) - 1:
            return csf

        c_turn = (tc <= tp + EPS)

        if c_turn:
            best = -1e100
            for i in range(p):
                if mask & (1 << i):
                    continue
                v = best_carlsson(mask | (1 << i),
                                  tc + areas[i], tp,
                                  csf + areas[i],
                                  alpha, beta)
                if v > best:
                    best = v
                if best > alpha:
                    alpha = best
                if alpha >= beta:
                    break
            return best
        else:
            best = 1e100
            for i in range(p):
                if mask & (1 << i):
                    continue
                v = best_carlsson(mask | (1 << i),
                                  tc, tp + areas[i],
                                  csf,
                                  alpha, beta)
                if v < best:
                    best = v
                if best < beta:
                    beta = best
                if alpha >= beta:
                    break
            return best

    total = sum(areas)
    carl = best_carlsson(0, 0.0, 0.0, 0.0, -1e100, 1e100)
    pooh = total - carl

    sys.stdout.write(f"{carl:.10f} {pooh:.10f}\n")

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

Notes:
- This Python version mirrors the C++ logic closely.
- With N ≤ 4, p is small and recursion is fine; Python should still be okay for typical constraints, but the original problem has a very tight 0.25s limit (meant for C++). For Python, performance may be borderline on worst cases.

---

## 5) Compressed editorial

1. Build all pieces created by N ≤ 4 lines inside a radius-R disk.
   - Vertices = all line–circle intersections + all line–line intersections inside disk (dedup).
   - Edges = split each line into segments between consecutive vertices on that line, plus circle boundary arcs between consecutive boundary vertices (sorted by angle).
   - Use half-edges with twins; at each vertex sort outgoing half-edges by direction angle; set `next(h)` to the edge immediately clockwise from `twin(h)` to trace faces.

2. Compute each face area by walking `next` cycles:
   - Segment contribution: `cross(P,Q)`
   - Arc contribution: `r^2 * dtheta` (signed, respecting CW/CCW)
   - Keep faces with positive signed area → pizza pieces.

3. Game:
   - `tc,tp` are cumulative eaten areas (times).
   - Next mover is Carlsson if `tc <= tp`, else Pooh.
   - Minimax over subsets of pieces: Carlsson maximizes his final sum; Pooh minimizes it.
   - Alpha-beta pruning; pieces ≤ ~11 so brute force works.

4. Output Carlsson and Pooh areas (`pooh = total - carlsson`).