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>
  void sc(int &x) { scanf("%d", &x); }
  void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
  void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
  void sc(long long &x) { scanf("%lld", &x); }
  void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
  void sc(long long &x, long long &y, long long &z) {
    scanf("%lld%lld%lld", &x, &y, &z);
  }
  void sc(char *x) { scanf("%s", x); }
  void sc(char *x, char *y) { scanf("%s%s", x, y); }
  void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
  void out(int x) { printf("%d\n", x); }
  void out(long long x) { printf("%lld\n", x); }
  void out(int x, int y) { printf("%d %d\n", x, y); }
  void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
  void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
  void out(long long x, long long y, long long z) {
    printf("%lld %lld %lld\n", x, y, z);
  }
  void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
  using namespace std;
  const int N = 205;
  mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
  int rdint(int l, int r) { return uniform_int_distribution<int>(l, r)(gen); }
  int n, k, x[N], y[N];
  pair<int, long long> dp[N][N];
  int cal(long long m) {
    for (int len = 1; len < n; len++)
      for (int l = 0; l + len < n; l++) {
        int r = l + len;
        if (r == l + 1)
          dp[l][r] = {0, 0};
        else {
          dp[l][r] = {0, 0};
          for (int k = l + 1; k < r; k++) {
            long long s = abs(1ll * (y[r] - y[l]) * (x[k] - x[l]) -
                              1ll * (x[r] - x[l]) * (y[k] - y[l]));
            pair<int, long long> cur = {dp[l][k].first + dp[k][r].first,
                                        dp[l][k].second + dp[k][r].second + s};
            if (cur.second >= m) cur.first++, cur.second = 0;
            dp[l][r] = max(dp[l][r], cur);
          }
        }
      }
    return dp[0][n - 1].first;
  }
  void sol(int cas) {
    sc(n, k);
    for (int i = 0; i <= n - 1; i++) sc(x[i], y[i]);
    long long l = 1, r = 1e18, ans;
    while (l <= r) {
      long long m = (l + r) >> 1;
      if (cal(m) >= k + 1)
        ans = m, l = m + 1;
      else
        r = m - 1;
    }
    out(ans);
  }
  int main() {
    srand(time(0));
    int t = 1, cas = 0;
    while (t--) {
      sol(++cas);
    }
    return 0;
  }
  