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;
  int n, m;
  struct Point {
    int x, y;
    Point(int _x = 0, int _y = 0) { x = _x, y = _y; }
    friend Point operator-(const Point &a, const Point &b) {
      return Point(a.x - b.x, a.y - b.y);
    }
    friend long long operator%(const Point &a, const Point &b) {
      return 1LL * a.x * b.y - 1LL * a.y * b.x;
    }
  } p[404];
  long long Z;
  bool check(long long z) {
    static pair<int, long long> dp[404][404];
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n * 2 - 2; ++i) {
      long long w = (p[i + 1] - p[i]) % (p[i + 2] - p[i]);
      dp[i][i + 2] = w >= z ? make_pair(1, 0LL) : make_pair(0, w);
      if (dp[i][i + 2].first >= m) return true;
    }
    for (int len = 4; len <= n + 1; ++len) {
      for (int l = 1; l + len - 1 <= 2 * n; ++l) {
        int r = l + len - 1;
        dp[l][r] = make_pair(0, 0LL);
        for (int k = l + 1; k < r; ++k) {
          auto t = make_pair(
              dp[l][k].first + dp[k][r].first,
              dp[l][k].second + dp[k][r].second + (p[k] - p[l]) % (p[r] - p[l]));
          if (t.second >= z) ++t.first, t.second = 0;
          if (t.first >= m) return true;
          dp[l][r] = max(dp[l][r], t);
        }
      }
    }
    return false;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    ++m;
    for (int i = 1; i <= n; ++i) {
      cin >> p[i].x >> p[i].y;
    }
    for (int i = 1; i <= n; ++i) {
      p[i + n] = p[i];
    }
    long long l = 1, r = 1e17 + 233, ans = 0;
    while (l < r) {
      long long mid = (l + r) >> 1;
      if (check(mid)) {
        ans = mid, l = mid + 1;
      } else {
        r = mid;
      }
    }
    cout << ans << '\n';
    return 0;
  }
  