#include <bits/stdc++.h>
using namespace std;
inline long long readint() {
  long long x = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f ? -x : x;
}
const int maxn = 200 + 5;
int n, k, x[maxn], y[maxn];
long long S(int a, int b, int c) {
  return abs(1ll * (x[b] - x[a]) * (y[c] - y[a]) -
             1ll * (x[c] - x[a]) * (y[b] - y[a]));
}
int f[maxn][maxn];
long long g[maxn][maxn];
int main() {
  n = readint();
  k = readint();
  for (int i = 1; i <= n; i++) {
    x[i] = readint();
    y[i] = readint();
  }
  long long l = 0, r = 8e16;
  while (l < r) {
    long long mid = l + (r - l + 1) / 2;
    for (int i = n; i > 0; i--)
      for (int j = i + 1; j <= n; j++) {
        f[i][j] = 0;
        g[i][j] = 0;
        for (int k = i + 1; k < j; k++) g[i][j] += S(j, k, k - 1);
        auto upd = [i, j](int f2, long long g2) {
          if (f2 > f[i][j] || (f2 == f[i][j] && g2 > g[i][j])) {
            f[i][j] = f2;
            g[i][j] = g2;
          }
        };
        for (int k = i + 1; k < j; k++)
          if (S(i, j, k) + g[i][k] + g[k][j] >= mid)
            upd(f[i][k] + f[k][j] + 1, 0);
          else
            upd(f[i][k] + f[k][j], S(i, j, k) + g[i][k] + g[k][j]);
      }
    if (f[1][n] >= k + 1)
      l = mid;
    else
      r = mid - 1;
  }
  printf("%lld\n", r);
  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;
inline long long readint() {
  long long x = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f ? -x : x;
}
const int maxn = 200 + 5;
int n, k, x[maxn], y[maxn];
long long S(int a, int b, int c) {
  return abs(1ll * (x[b] - x[a]) * (y[c] - y[a]) -
             1ll * (x[c] - x[a]) * (y[b] - y[a]));
}
int f[maxn][maxn];
long long g[maxn][maxn];
int main() {
  n = readint();
  k = readint();
  for (int i = 1; i <= n; i++) {
    x[i] = readint();
    y[i] = readint();
  }
  long long l = 0, r = 8e16;
  while (l < r) {
    long long mid = l + (r - l + 1) / 2;
    for (int i = n; i > 0; i--)
      for (int j = i + 1; j <= n; j++) {
        f[i][j] = 0;
        g[i][j] = 0;
        for (int k = i + 1; k < j; k++) g[i][j] += S(j, k, k - 1);
        auto upd = [i, j](int f2, long long g2) {
          if (f2 > f[i][j] || (f2 == f[i][j] && g2 > g[i][j])) {
            f[i][j] = f2;
            g[i][j] = g2;
          }
        };
        for (int k = i + 1; k < j; k++)
          if (S(i, j, k) + g[i][k] + g[k][j] >= mid)
            upd(f[i][k] + f[k][j] + 1, 0);
          else
            upd(f[i][k] + f[k][j], S(i, j, k) + g[i][k] + g[k][j]);
      }
    if (f[1][n] >= k + 1)
      l = mid;
    else
      r = mid - 1;
  }
  printf("%lld\n", r);
  return 0;
}

```