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 = 204;
  const long long INF = 1000000000000000000;
  int n, k;
  struct data {
    int ans;
    long long S;
    void chkmax(data x) {
      if (x.ans > ans || (x.ans == ans && x.S > S)) ans = x.ans, S = x.S;
    }
  } f[N][N];
  struct point {
    long long x, y;
  } p[N];
  point operator-(point a, point b) { return (point){a.x - b.x, a.y - b.y}; }
  long long operator*(point a, point b) { return a.x * b.y - a.y * b.x; }
  data operator+(data x, data y) { return (data){x.ans + y.ans, x.S + y.S}; }
  bool check(long long S) {
    for (int j = 3; j <= n; ++j)
      for (int i = j - 2; i >= 1; --i) {
        f[i][j] = (data){0, 0};
        for (int k = i; k <= j; ++k) {
          data tmp = f[i][k] + f[k][j];
          tmp.S += abs((p[i] - p[k]) * (p[j] - p[k]));
          if (tmp.S >= S) tmp.S = 0, ++tmp.ans;
          f[i][j].chkmax(tmp);
        }
      }
    return f[1][n].ans > k;
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i) scanf("%lld%lld", &p[i].x, &p[i].y);
    long long l = 1, r = INF;
    while (l <= r)
      check((l + r >> 1)) ? l = (l + r >> 1) + 1 : r = (l + r >> 1) - 1;
    printf("%lld\n", l - 1);
    return 0;
  }
  