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;
  using LL = long long;
  const int N = 205;
  int n, K, x[N], y[N];
  struct node {
    LL area;
    int cnt;
    bool operator<(const node& k) const {
      return cnt != k.cnt ? cnt < k.cnt : area < k.area;
    }
  } f[N][N];
  LL cross(LL a, LL b, LL c, LL d) { return a * d - b * c; }
  bool chk(LL mid) {
    for (int i = (1); i <= (n); ++i)
      for (int j = (i + 2); j <= (n); ++j) {
        f[i][j] = (node){(LL)1e18, (int)-1e9};
      }
    for (int len = (3); len <= (n); ++len) {
      for (int i = (1); i <= (n); ++i) {
        int j = i + len - 1;
        if (j > n) break;
        for (int k = (i + 1); k <= (j - 1); ++k) {
          node cur = (node){
              f[i][k].area + f[k][j].area +
                  cross(x[k] - x[i], y[k] - y[i], x[j] - x[i], y[j] - y[i]),
              f[i][k].cnt + f[k][j].cnt};
          if (cur.area >= mid) {
            cur.area = 0, ++cur.cnt;
          }
          f[i][j] = max(f[i][j], cur);
        }
      }
    }
    return f[1][n].cnt >= K;
  }
  int main() {
    scanf("%d%d", &n, &K), ++K;
    for (int i = (1); i <= (n); ++i) scanf("%d%d", &x[i], &y[i]);
    LL l = 0, r = 1e18, ans = 0;
    chk(11);
    while (l <= r) {
      LL mid = (l + r) >> 1;
      if (chk(mid))
        ans = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    printf("%lld\n", ans);
    return 0;
  }
  