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;
  inline long long read() {
    char ch = getchar();
    long long nega = 1;
    while (!isdigit(ch)) {
      if (ch == '-') nega = -1;
      ch = getchar();
    }
    long long ans = 0;
    while (isdigit(ch)) {
      ans = ans * 10 + ch - 48;
      ch = getchar();
    }
    if (nega == -1) return -ans;
    return ans;
  }
  void print(vector<long long> x) {
    for (long long i = 0; i < (long long)x.size(); i++)
      printf("%lld%c", x[i], " \n"[i == (long long)x.size() - 1]);
  }
  struct Node {
    long long v, r;
  };
  Node max(Node a, Node b) {
    if (a.v > b.v || (a.v == b.v && a.r > b.r))
      return a;
    else
      return b;
  }
  struct Vec {
    long long x, y;
  } a[205];
  Vec operator-(Vec a, Vec b) { return (Vec){a.x - b.x, a.y - b.y}; }
  long long cross(Vec a, Vec b) { return a.x * b.y - a.y * b.x; }
  Node f[205][205];
  long long n, m;
  bool chk(long long mid) {
    for (long long i = 1; i <= n; i++)
      for (long long j = i + 2; j <= n; j++)
        f[i][j] = (Node){-0x3f3f3f3f, -0x3f3f3f3f};
    for (long long len = 3; len <= n; len++) {
      for (long long i = 1; i + len - 1 <= n; i++) {
        long long j = i + len - 1;
        for (long long k = i + 1; k <= j; k++) {
          Node go =
              (Node){f[i][k].v + f[k][j].v,
                     f[i][k].r + f[k][j].r + cross(a[i] - a[j], a[k] - a[i])};
          if (go.r >= mid) go.v++, go.r = 0;
          f[i][j] = max(f[i][j], go);
        }
      }
    }
    return f[1][n].v >= m;
  }
  signed main() {
    cin >> n >> m;
    m++;
    for (long long i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
    long long l = 0, r = 0x3f3f3f3f3f3f3f3f, ans = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (chk(mid))
        l = mid + 1, ans = mid;
      else
        r = mid - 1;
    }
    cout << ans << endl;
    return 0;
  }
  