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 = 1e6 + 10;
  const long long mod = 1e9 + 7;
  inline long long gcd(long long a, long long b) {
    return !b ? a : gcd(b, a % b);
  }
  inline long long q_pow(long long a, long long x = mod - 2) {
    long long ans = 1, tmp = a;
    while (x) {
      if (x & 1) (ans *= tmp) %= mod;
      (tmp *= tmp) %= mod;
      x >>= 1;
    }
    return ans;
  }
  template <typename T>
  inline void re(T &N) {
    int f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9')
      if (c == '-') f = -1;
    N = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
    N *= f;
  }
  template <class T, class... T_>
  inline void re(T &x, T_ &...y) {
    re(x), re(y...);
  }
  int m, n, t = 1, st, en;
  pair<long long, long long> a[N];
  pair<long long, long long> dp[210][210];
  pair<long long, long long> dfs(int l, int r, long long mid) {
    if (l + 1 == r) return {0, 0};
    if (~dp[l][r].first) return dp[l][r];
    auto &ans = dp[l][r] = {0, 0};
    for (int i = l + 1; i < r; i++) {
      auto dl = dfs(l, i, mid);
      auto dr = dfs(i, r, mid);
      auto area = [&](int i, int j, int k) {
        auto det = [&](int i, int j) {
          return a[i].first * a[j].second - a[i].second * a[j].first;
        };
        return det(i, j) + det(j, k) + det(k, i);
      };
      pair<long long, long long> tmp = {dl.first + dr.first,
                                        dl.second + dr.second + area(l, i, r)};
      if (tmp.second >= mid) tmp.first++, tmp.second = 0;
      ans = max(ans, tmp);
    }
    return ans;
  }
  int main() {
    re(n, m);
    for (int i = 1; i <= n; i++) re(a[i].first, a[i].second);
    long long l = 0, r = 2e17, ans = l;
    while (l <= r) {
      auto check = [&](long long mid) {
        memset(dp, -1, sizeof(dp));
        pair<long long, long long> ans = dfs(1, n, mid);
        return ans.first > m;
      };
      long long mid = (l + r) >> 1;
      if (check(mid))
        l = mid + 1, ans = mid;
      else
        r = mid - 1;
    }
    printf("%lld\n", ans);
  }
  