#include <bits/stdc++.h>
using namespace std;
const int maxn = 210;
int n, K;
long long x[maxn], y[maxn];
pair<int, long long> f[maxn][maxn];
int main() {
  scanf("%d %d", &n, &K), K++;
  for (int i = 1; i <= n; i++) {
    scanf("%lld %lld", &x[i], &y[i]);
  }
  auto cross = [&](int i, int j, int k) {
    long long x1 = x[j] - x[i], y1 = y[j] - y[i];
    long long x2 = x[k] - x[i], y2 = y[k] - y[i];
    return x1 * y2 - x2 * y1;
  };
  auto chk = [&](long long x) {
    for (int len = 3; len <= n; len++) {
      for (int l = 1, r = len; r <= n; l++, r++) {
        f[l][r] = {-1e9, 0};
        for (int k = l + 1; k < r; k++) {
          int a = f[l][k].first + f[k][r].first;
          long long b = f[l][k].second + f[k][r].second + cross(l, k, r);
          if (b >= x) a++, b = 0;
          f[l][r] = max(f[l][r], make_pair(a, b));
        }
      }
    }
    return f[1][n].first >= K;
  };
  long long l = 0, r = 1e17, ans;
  while (l <= r) {
    long long mid = (l + r) / 2;
    chk(mid) ? l = (ans = mid) + 1 : r = mid - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 210;
int n, K;
long long x[maxn], y[maxn];
pair<int, long long> f[maxn][maxn];
int main() {
  scanf("%d %d", &n, &K), K++;
  for (int i = 1; i <= n; i++) {
    scanf("%lld %lld", &x[i], &y[i]);
  }
  auto cross = [&](int i, int j, int k) {
    long long x1 = x[j] - x[i], y1 = y[j] - y[i];
    long long x2 = x[k] - x[i], y2 = y[k] - y[i];
    return x1 * y2 - x2 * y1;
  };
  auto chk = [&](long long x) {
    for (int len = 3; len <= n; len++) {
      for (int l = 1, r = len; r <= n; l++, r++) {
        f[l][r] = {-1e9, 0};
        for (int k = l + 1; k < r; k++) {
          int a = f[l][k].first + f[k][r].first;
          long long b = f[l][k].second + f[k][r].second + cross(l, k, r);
          if (b >= x) a++, b = 0;
          f[l][r] = max(f[l][r], make_pair(a, b));
        }
      }
    }
    return f[1][n].first >= K;
  };
  long long l = 0, r = 1e17, ans;
  while (l <= r) {
    long long mid = (l + r) / 2;
    chk(mid) ? l = (ans = mid) + 1 : r = mid - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```