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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long x[205], y[205];
  long long area[205][205][205];
  pair<long long, long long> dp[205][205];
  long long n, cuts;
  bool check(long long s) {
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= n; j++) dp[i][j] = make_pair(0LL, 0LL);
    }
    for (long long d = 2; d < n; d++) {
      for (long long i = 1; i <= n; i++) {
        long long j = (i + d - 1) % n + 1;
        for (long long jump = 1; jump < d; jump++) {
          long long k = (i + jump - 1) % n + 1;
          long long cnt = dp[i][k].first + dp[k][j].first,
                    spare = dp[i][k].second + dp[k][j].second + area[i][j][k];
          if (spare >= s) {
            cnt++;
            spare = 0;
          }
          dp[i][j] = max(dp[i][j], make_pair(cnt, spare));
        }
      }
    }
    for (long long i = 1; i <= n; i++) {
      if (dp[i][(i + n - 2) % n + 1].first >= cuts + 1) return 1;
    }
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> cuts;
    for (long long i = 1; i <= n; i++) cin >> x[i] >> y[i];
    for (long long i = 1; i <= n; i++) {
      for (long long j = i + 1; j <= n; j++) {
        for (long long k = j + 1; k <= n; k++) {
          area[i][j][k] = x[i] * y[j] + x[j] * y[k] + x[k] * y[i] - x[j] * y[i] -
                          x[k] * y[j] - x[i] * y[k];
          if (area[i][j][k] < 0) area[i][j][k] = -area[i][j][k];
          area[i][k][j] = area[j][i][k] = area[j][k][i] = area[k][i][j] =
              area[k][j][i] = area[i][j][k];
        }
      }
    }
    long long l = 1, r = 8e16;
    while (l < r) {
      long long mid = (l + r + 1) / 2;
      if (check(mid))
        l = mid;
      else
        r = mid - 1;
    }
    cout << l << "\n";
  }
  