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 _Tp>
  void read(_Tp &x) {
    char ch(getchar());
    bool f(false);
    while (!isdigit(ch)) f |= ch == 45, ch = getchar();
    x = ch & 15, ch = getchar();
    while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
    if (f) x = -x;
  }
  template <typename _Tp, typename... Args>
  void read(_Tp &t, Args &...args) {
    read(t);
    read(args...);
  }
  template <typename _Tp>
  inline void chmax(_Tp &a, const _Tp &b) {
    a = a < b ? b : a;
  }
  const int N = 205;
  int n, k;
  int x[N], y[N], dp[N][N];
  long long a[N][N], all;
  inline int nxt(int x) { return x + 1 == n ? 0 : x + 1; }
  inline int pre(int x) { return !x ? n - 1 : x - 1; }
  inline long long cross(long long x1, long long y1, long long x2, long long y2) {
    return x1 * y2 - x2 * y1;
  }
  bool check(long long mid) {
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) dp[i][j] = 0;
    for (int len = 3; len <= n; ++len) {
      int i = 0, j = len - 1;
      do {
        if (a[i][j] < mid) {
          i = nxt(i), j = nxt(j);
          continue;
        }
        dp[i][j] = 1;
        for (int k = nxt(i), p = i; k != j; k = nxt(k)) {
          while (a[k][p] - a[j][i] < mid) p = nxt(p);
          chmax(dp[i][j], dp[p][k] + 1);
        }
        for (int k = nxt(i), p = k; k != j; k = nxt(k)) {
          while (p < j && a[p][k] - a[j][i] - a[p][j] - a[i][k] < mid) p = nxt(p);
          if (a[p][k] - a[j][i] - a[p][j] - a[i][k] < mid) break;
          chmax(dp[i][j], dp[p][j] + dp[i][k] + 1);
        }
        i = nxt(i), j = nxt(j);
      } while (i);
    }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (dp[i][j] + dp[j][i] >= k + 1) return true;
    return false;
  }
  int main() {
    read(n, k);
    for (int i = 0; i < n; ++i) read(x[i], y[i]);
    for (int i = 0; i < n; ++i) {
      int j = nxt(i), k;
      long long s = 0;
      for (; j != i; j = k)
        a[i][j] = s, k = nxt(j),
        s += cross(x[j] - x[i], y[j] - y[i], x[k] - x[i], y[k] - y[i]);
    }
    all = a[0][n - 1];
    for (int i = 0; i < n; ++i) a[i][i] = all;
    long long l = 0, r = 1e18, mid;
    while (l < r) mid = (l + r + 1) >> 1, check(mid) ? l = mid : r = mid - 1;
    printf("%lld\n", l);
    return 0;
  }
  