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;
  const long long INF = 1LL << 61LL;
  struct point {
    int x, y;
    point(int _x = 0, int _y = 0) {
      x = _x;
      y = _y;
    }
    point operator-(const point z) const { return point(x - z.x, y - z.y); }
  } a[202];
  struct val {
    long long num;
    long long remain;
    val(long long _a = 0, long long _b = 0) {
      num = _a;
      remain = _b;
    }
    val operator+(const val x) const {
      return val(num + x.num, remain + x.remain);
    }
    bool operator<(const val x) const {
      if (x.num == num) return remain < x.remain;
      return num < x.num;
    }
  } f[202][202], tmp;
  int n, m;
  long long cross(point a, point b) {
    return abs(1LL * a.x * b.y - 1LL * a.y * b.x);
  }
  bool chk(long long x) {
    int i, j, k, len;
    for (i = 1; i <= n; ++i) {
      for (j = i + 2; j <= n; ++j) f[i][j] = val(-(INF >> 1), -(INF >> 1));
    }
    for (len = 2; len < n; ++len) {
      for (i = 1; i + len <= n; ++i) {
        j = i + len;
        for (k = i + 1; k < j; ++k) {
          tmp = f[i][k] + f[k][j];
          tmp.remain += cross(a[k] - a[j], a[i] - a[j]);
          if (tmp.remain >= x) {
            ++tmp.num;
            tmp.remain = 0;
          }
          if (f[i][j] < tmp) f[i][j] = tmp;
        }
      }
    }
    return f[1][n].num >= m;
  }
  int main() {
    int i, j;
    long long l = 0, r = INF, mid, ans = -1;
    scanf("%d%d", &n, &m);
    ++m;
    for (i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
    while (l <= r) {
      mid = (l + r) >> 1;
      if (chk(mid)) {
        ans = mid;
        l = mid + 1;
      } else
        r = mid - 1;
    }
    printf("%lld\n", ans);
    return 0;
  }
  