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() {
    long long x = 0, f = 1;
    char c = getchar();
    while ((c < '0' || c > '9') && (c != '-')) c = getchar();
    if (c == '-') f = -1, c = getchar();
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
  }
  const int N = 410;
  int n, m;
  pair<int, int> p[N];
  pair<long long, long long> f[N][N];
  inline pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
    return make_pair(a.first - b.first, a.second - b.second);
  }
  inline long long cross(pair<int, int> a, pair<int, int> b) {
    return 1ll * a.first * b.second - 1ll * a.second * b.first;
  }
  inline long long S(pair<int, int> x, pair<int, int> y, pair<int, int> z) {
    return abs(cross(y - x, z - x));
  }
  inline bool check(long long x) {
    for (int i = (1); i <= (n); i++)
      for (int j = (i + 2); j <= (i + n - 1); j++) f[i][j] = make_pair(0, 0);
    for (int len = (3); len <= (n); len++) {
      for (int i = (1); i <= (n); i++) {
        int j = i + len - 1;
        for (int k = (i + 1); k <= (j - 1); k++) {
          int a = f[i][k].first + f[k][j].first;
          long long b = f[i][k].second + f[k][j].second + S(p[i], p[k], p[j]);
          if (b >= x) b = 0, ++a;
          if (a > f[i][j].first || (a == f[i][j].first && b > f[i][j].second))
            f[i][j] = make_pair(a, b);
        }
      }
    }
    for (int i = (1); i <= (n); i++)
      if (f[i][i + n - 1].first >= m + 1) return 1;
    return 0;
  }
  int main() {
    n = read(), m = read();
    for (int i = (1); i <= (n); i++)
      p[i].first = read(), p[i].second = read(), p[i + n] = p[i];
    long long l = 1, r = 8e16, ret = 0;
    while (l <= r) {
      long long mid = l + r >> 1;
      if (check(mid))
        ret = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    printf("%lld\n", ret);
  }
  