description: You are given a strictly convex polygon with n vertices.
  
  You will make k cuts that meet the following conditions: 
  
    * each cut is a segment that connects two different nonadjacent vertices; 
    * two cuts can intersect only at vertices of the polygon. 
  
  
  
  Your task is to maximize the area of the smallest region that will be formed by the polygon and those k cuts.
  
  Input
  
  The first line contains two integers n, k (3 ≤ n ≤ 200, 0 ≤ k ≤ n-3).
  
  The following n lines describe vertices of the polygon in anticlockwise direction. The i-th line contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^8) — the coordinates of the i-th vertex.
  
  It is guaranteed that the polygon is convex and that no two adjacent sides are parallel.
  
  Output
  
  Print one integer: the maximum possible area of the smallest region after making k cuts multiplied by 2.
  
  Examples
  
  Input
  
  
  8 4
  -2 -4
  2 -2
  4 2
  1 5
  0 5
  -4 4
  -5 0
  -5 -1
  
  
  Output
  
  
  11
  
  
  Input
  
  
  6 3
  2 -2
  2 -1
  1 2
  0 2
  -2 1
  -1 0
  
  
  Output
  
  
  3
  
  Note
  
  In the first example, it's optimal to make cuts between the following pairs of vertices: 
  
    * (-2, -4) and (4, 2), 
    * (-2, -4) and (1, 5), 
    * (-5, -1) and (1, 5), 
    * (-5, 0) and (0, 5). 
  
  <image> Points (-5, -1), (1, 5), (0, 5), (-5, 0) determine the smallest region with double area of 11. 
  
  In the second example, it's optimal to make cuts between the following pairs of vertices: 
  
    * (2, -1) and (0, 2), 
    * (2, -1) and (1, 0), 
    * (-1, 0) and (0, 2). 
  
  <image> Points (2, -2), (2, -1), (-1, 0) determine one of the smallest regions with double area of 3. 
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 205;
  struct point {
    int x, y;
    point() {}
    point(int _x, int _y) { x = _x, y = _y; }
    point operator-(const point& b) const { return point(x - b.x, y - b.y); }
  };
  pair<long long, long long> add(pair<long long, long long> a,
                                 pair<long long, long long> b) {
    return {a.first + b.first, a.second + b.second};
  }
  long long cross(point a, point b) { return 1ll * a.x * b.y - 1ll * a.y * b.x; }
  point a[N];
  pair<long long, long long> f[N][N];
  int n, K;
  bool check(long long x) {
    for (int len = 3; len <= n; ++len)
      for (int l = 1; l + len - 1 <= n; ++l) {
        int r = l + len - 1;
        f[l][r] = {-1e9, 0};
        for (int k = l + 1; k < r; ++k) {
          pair<long long, long long> c = add(f[l][k], f[k][r]);
          c.second += cross(a[k] - a[l], a[r] - a[l]);
          if (c.second >= x) c.second = 0, c.first++;
          f[l][r] = max(f[l][r], c);
        }
      }
    return f[1][n].first >= K;
  }
  signed main() {
    cin >> n >> K;
    ++K;
    for (int i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y;
    long long l = 0, r = 1e18;
    while (l < r) {
      long long mid = (l + r + 1) >> 1;
      if (check(mid))
        l = mid;
      else
        r = mid - 1;
    }
    cout << l;
  }
  