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;
  using ll = long long;
  struct P {
    ll x, y;
    P(ll xx = 0, ll yy = 0) { x = xx, y = yy; }
    P operator-(P r) { return P(x - r.x, y - r.y); }
  } p[205];
  ll Cross(P A, P B) { return A.x * B.y - A.y * B.x; }
  pair<int, ll> dp[205][205];
  int n, t;
  int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> t;
    for (int i = 1; i <= n; i += 1) {
      cin >> p[i].x >> p[i].y;
    }
    ll lo = -1e18, hi = 1e18, ans = -1e18;
    while (hi >= lo) {
      ll mid = lo + hi >> 1;
      for (int len = 2; len <= n; len += 1) {
        for (int l = 1; l + len - 1 <= n; l += 1) {
          int r = l + len - 1;
          if (len == 2)
            dp[l][r] = make_pair(0, 0);
          else {
            dp[l][r] = make_pair(-1000000000, 0);
            for (int k = l + 1; k < r; k += 1) {
              int nk = dp[l][k].first + dp[k][r].first;
              ll area = dp[l][k].second + dp[k][r].second +
                        abs(Cross(p[l] - p[r], p[k] - p[r]));
              if (area >= mid) {
                area = 0;
                nk += 1;
              }
              dp[l][r] = max(dp[l][r], make_pair(nk, area));
            }
          }
        }
      }
      if (dp[1][n].first >= t + 1) {
        lo = mid + 1;
        ans = mid;
      } else
        hi = mid - 1;
    }
    cout << ans << endl;
    return 0;
  }
  