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>
  #pragma GCC optimize(2, 3, "Ofast")
  #pragma GCC target("avx", "avx2")
  using namespace std;
  template <typename T1, typename T2>
  void ckmin(T1 &a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  void ckmax(T1 &a, T2 b) {
    if (a < b) a = b;
  }
  int read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void print(T x, char let) {
    print(x), putchar(let);
  }
  const int N = 205;
  template <typename T>
  int sign(T x) {
    if (x < 0) return -1;
    if (x == 0) return 0;
    return 1;
  }
  struct Vec {
    long long x, y;
    Vec(long long _x = 0, long long _y = 0) { x = _x, y = _y; }
  } a[N];
  Vec operator+(Vec x, Vec y) { return Vec(x.x + y.x, x.y + y.y); }
  Vec operator-(Vec x, Vec y) { return Vec(x.x - y.x, x.y - y.y); }
  long long Cross(Vec x, Vec y) { return x.x * y.y - x.y * y.x; }
  int n, m;
  pair<long long, long long> dp[N][N];
  bool check(long long area) {
    for (int len = 3; len <= n; len++) {
      for (int i = 1, j = len; j <= n; i++, j++) {
        dp[i][j] = {-1e9, 0};
        for (int k = i + 1; k <= j - 1; k++) {
          pair<long long, long long> f = {dp[i][k].first + dp[k][j].first,
                                          dp[i][k].second + dp[k][j].second +
                                              Cross(a[k] - a[i], a[j] - a[i])};
          if (f.second >= area) f.first++, f.second = 0;
          ckmax(dp[i][j], f);
        }
      }
    }
    return dp[1][n].first >= m;
  }
  int main() {
    n = read(), m = read() + 1;
    for (int i = 1; i <= n; i++) a[i].x = read(), a[i].y = read();
    long long l = 0, r = 1e18;
    while (l < r) {
      long long mid = l + r + 1 >> 1;
      if (check(mid))
        l = mid;
      else
        r = mid - 1;
    }
    print(l, '\n');
    return 0;
  }
  