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 (!isdigit(c)) {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = (x << 3) + (x << 1) + (c ^ 48);
      c = getchar();
    }
    return f == -1 ? ~x + 1 : x;
  }
  long long n, k;
  struct node {
    long long x, y;
  } a[210];
  pair<long long, long long> dp[210][210];
  node operator-(node x, node y) { return node{x.x - y.x, x.y - y.y}; }
  long long operator*(node x, node y) { return x.x * y.y - x.y * y.x; }
  long long get(long long i, long long j, long long k) {
    return (a[j] - a[i]) * (a[k] - a[i]);
  }
  long long check(long long mid) {
    for (long long i = 1; i <= n; ++i) {
      for (long long j = i + 2; j <= n; ++j) dp[i][j] = make_pair(-1e8, -1e8);
    }
    for (long long l = 3; l <= n; ++l) {
      for (long long i = 1; i + l - 1 <= n; ++i) {
        long long j = i + l - 1;
        for (long long k = i; k <= j; ++k) {
          pair<long long, long long> nw =
              make_pair(dp[i][k].first + dp[k][j].first,
                        dp[i][k].second + dp[k][j].second + get(i, k, j));
          if (nw.second >= mid) ++nw.first, nw.second = 0;
          dp[i][j] = max(dp[i][j], nw);
        }
      }
    }
    return dp[1][n].first >= k + 1;
  }
  signed main() {
    n = read(), k = read();
    for (long long i = 1; i <= n; ++i) a[i].x = read(), a[i].y = read();
    long long l = 0, r = 1e18, ans = 0;
    while (l <= r) {
      long long mid = l + r >> 1;
      if (check(mid))
        ans = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    printf("%lld\n", ans);
  }
  