#include <bits/stdc++.h>
using namespace std;
int dp[205][205];
long long rest[205][205];
int n, x[205], y[205], m;
long long area(int x1, int y1, int x2, int y2, int x3, int y3) {
  long long a = x2 - x1, b = y2 - y1, c = x3 - x1, d = y3 - y1;
  return abs(a * d - b * c);
}
bool chk(long long X) {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = -1e9;
  for (int i = 1; i < n; i++) dp[i][i + 1] = 0, rest[i][i + 1] = 0;
  for (int le = 3; le <= n; le++)
    for (int l = 1, r = le; r <= n; l++, r++) {
      for (int k = l + 1; k < r; k++) {
        int cnt = dp[l][k] + dp[k][r];
        long long S =
            area(x[l], y[l], x[r], y[r], x[k], y[k]) + rest[l][k] + rest[k][r];
        if (S >= X) S = 0, cnt++;
        if (cnt > dp[l][r])
          dp[l][r] = cnt, rest[l][r] = S;
        else if (cnt == dp[l][r])
          rest[l][r] = max(rest[l][r], S);
      }
    }
  return dp[1][n] >= m;
}
int main() {
  scanf("%d%d", &n, &m);
  m++;
  for (int i = 1; i <= n; i++) scanf("%d%d", x + i, y + i);
  long long l = 0, r = 8e16, ans = 0;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (chk(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", ans);
}

### 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;
int dp[205][205];
long long rest[205][205];
int n, x[205], y[205], m;
long long area(int x1, int y1, int x2, int y2, int x3, int y3) {
  long long a = x2 - x1, b = y2 - y1, c = x3 - x1, d = y3 - y1;
  return abs(a * d - b * c);
}
bool chk(long long X) {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = -1e9;
  for (int i = 1; i < n; i++) dp[i][i + 1] = 0, rest[i][i + 1] = 0;
  for (int le = 3; le <= n; le++)
    for (int l = 1, r = le; r <= n; l++, r++) {
      for (int k = l + 1; k < r; k++) {
        int cnt = dp[l][k] + dp[k][r];
        long long S =
            area(x[l], y[l], x[r], y[r], x[k], y[k]) + rest[l][k] + rest[k][r];
        if (S >= X) S = 0, cnt++;
        if (cnt > dp[l][r])
          dp[l][r] = cnt, rest[l][r] = S;
        else if (cnt == dp[l][r])
          rest[l][r] = max(rest[l][r], S);
      }
    }
  return dp[1][n] >= m;
}
int main() {
  scanf("%d%d", &n, &m);
  m++;
  for (int i = 1; i <= n; i++) scanf("%d%d", x + i, y + i);
  long long l = 0, r = 8e16, ans = 0;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (chk(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", ans);
}

```