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;
  int dp[205][205];
  long long rest[205][205];
  int n, x[205], y[205], m;
  long long area(int x1, int y1, int x2, int y2, int x3, int y3) {
    long long a = x2 - x1, b = y2 - y1, c = x3 - x1, d = y3 - y1;
    return abs(a * d - b * c);
  }
  bool chk(long long X) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) dp[i][j] = -1e9;
    for (int i = 1; i < n; i++) dp[i][i + 1] = 0, rest[i][i + 1] = 0;
    for (int le = 3; le <= n; le++)
      for (int l = 1, r = le; r <= n; l++, r++) {
        for (int k = l + 1; k < r; k++) {
          int cnt = dp[l][k] + dp[k][r];
          long long S =
              area(x[l], y[l], x[r], y[r], x[k], y[k]) + rest[l][k] + rest[k][r];
          if (S >= X) S = 0, cnt++;
          if (cnt > dp[l][r])
            dp[l][r] = cnt, rest[l][r] = S;
          else if (cnt == dp[l][r])
            rest[l][r] = max(rest[l][r], S);
        }
      }
    return dp[1][n] >= m;
  }
  int main() {
    scanf("%d%d", &n, &m);
    m++;
    for (int i = 1; i <= n; i++) scanf("%d%d", x + i, y + i);
    long long l = 0, r = 8e16, ans = 0;
    while (l <= r) {
      long long mid = l + r >> 1;
      if (chk(mid))
        ans = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    printf("%lld\n", ans);
  }
  