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>
  template <typename T>
  bool chkmax(T &x, T y) {
    return x < y ? x = y, true : false;
  }
  template <typename T>
  bool chkmin(T &x, T y) {
    return x > y ? x = y, true : false;
  }
  using namespace std;
  long long ksm(long long a, long long b) {
    if (!b) return 1;
    long long ns = ksm(a, b >> 1);
    ns = ns * ns % 998244353;
    if (b & 1) ns = ns * a % 998244353;
    return ns;
  }
  pair<int, long long> dp[205][205];
  long long x[205], y[205], S;
  int nxt[205], n, k;
  int check(long long s) {
    S = s;
    int d = 0;
    for (int dis = 1; dis < n; dis++) {
      for (int r = 0; r < n; r++) {
        int l = (r + dis) % n;
        if (dis == 1) {
          dp[l][r] = make_pair(0, 0);
          continue;
        }
        dp[l][r] = make_pair(-1, -1);
        for (int k = nxt[r]; k != l; k = nxt[k]) {
          long long s =
              (x[r] - x[l]) * (y[k] - y[l]) - (x[k] - x[l]) * (y[r] - y[l]);
          auto s1 = dp[l][k], s2 = dp[k][r];
          int p1 = s1.first + s2.first;
          long long p2 = s1.second + s2.second + s;
          if (p2 >= S) p2 = 0, p1++;
          dp[l][r] = max(dp[l][r], make_pair(p1, p2));
        }
      }
    }
    for (int i = 0; i < n; i++) {
      d = max(d, dp[i][nxt[i]].first);
    }
    return d;
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++)
      scanf("%lld%lld", x + i, y + i), nxt[i] = (i + 1) % n;
    long long l = 0, r = (1ll << 60);
    while (l + 1 < r) {
      long long mid = (l + r) / 2;
      if (check(mid) >= k + 1)
        l = mid;
      else
        r = mid;
    }
    printf("%lld\n", l);
  }
  