#include <bits/stdc++.h>
using namespace std;
struct pnt {
  long long x, y;
  pnt(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
} P[300];
long long operator*(pnt A, pnt B) { return A.x * B.y - A.y * B.x; }
long long area(pnt A, pnt B, pnt C) { return A * B + B * C + C * A; }
pair<int, long long> dp[300][300];
long long sz[300][300];
int main() {
  int n = 0, k = 0;
  scanf("%d%d", &n, &k);
  k++;
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &P[i].x, &P[i].y);
  for (int l = 1; l <= n; l++) {
    for (int r = l + 1; r <= n; r++) sz[l][r] = sz[l][r - 1] + P[r - 1] * P[r];
    for (int r = l + 1; r <= n; r++) sz[l][r] += P[r] * P[l];
  }
  long long S = 0;
  for (int i = 1; i < n; i++) S += P[i] * P[i + 1];
  S += P[n] * P[1];
  long long L = 0, R = S / k;
  while (L < R) {
    long long x = (L + R + 1) >> 1;
    memset(dp, 0, sizeof(dp));
    for (int len = 3; len <= n; len++) {
      for (int l = 1; l + len - 1 <= n; l++) {
        int r = l + len - 1;
        if (sz[l][r] < x) continue;
        dp[l][r] = make_pair(1, 0);
        for (int k = l + 1; k <= r - 1; k++) {
          pair<int, long long> val = make_pair(
              dp[l][k].first + dp[k][r].first,
              dp[l][k].second + dp[k][r].second + area(P[l], P[k], P[r]));
          if (val.second >= x) val.first++, val.second = 0;
          dp[l][r] = max(dp[l][r], val);
        }
      }
    }
    if (dp[1][n].first >= k)
      L = x;
    else
      R = x - 1;
  }
  printf("%lld", L);
}

### Prompt

In cpp, your task is to solve 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;
struct pnt {
  long long x, y;
  pnt(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
} P[300];
long long operator*(pnt A, pnt B) { return A.x * B.y - A.y * B.x; }
long long area(pnt A, pnt B, pnt C) { return A * B + B * C + C * A; }
pair<int, long long> dp[300][300];
long long sz[300][300];
int main() {
  int n = 0, k = 0;
  scanf("%d%d", &n, &k);
  k++;
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &P[i].x, &P[i].y);
  for (int l = 1; l <= n; l++) {
    for (int r = l + 1; r <= n; r++) sz[l][r] = sz[l][r - 1] + P[r - 1] * P[r];
    for (int r = l + 1; r <= n; r++) sz[l][r] += P[r] * P[l];
  }
  long long S = 0;
  for (int i = 1; i < n; i++) S += P[i] * P[i + 1];
  S += P[n] * P[1];
  long long L = 0, R = S / k;
  while (L < R) {
    long long x = (L + R + 1) >> 1;
    memset(dp, 0, sizeof(dp));
    for (int len = 3; len <= n; len++) {
      for (int l = 1; l + len - 1 <= n; l++) {
        int r = l + len - 1;
        if (sz[l][r] < x) continue;
        dp[l][r] = make_pair(1, 0);
        for (int k = l + 1; k <= r - 1; k++) {
          pair<int, long long> val = make_pair(
              dp[l][k].first + dp[k][r].first,
              dp[l][k].second + dp[k][r].second + area(P[l], P[k], P[r]));
          if (val.second >= x) val.first++, val.second = 0;
          dp[l][r] = max(dp[l][r], val);
        }
      }
    }
    if (dp[1][n].first >= k)
      L = x;
    else
      R = x - 1;
  }
  printf("%lld", L);
}

```