#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 205;
int n, K, x[N], y[N];
struct node {
  LL area;
  int cnt;
  bool operator<(const node& k) const {
    return cnt != k.cnt ? cnt < k.cnt : area < k.area;
  }
} f[N][N];
LL cross(LL a, LL b, LL c, LL d) { return a * d - b * c; }
bool chk(LL mid) {
  for (int i = (1); i <= (n); ++i)
    for (int j = (i + 2); j <= (n); ++j) {
      f[i][j] = (node){(LL)1e18, (int)-1e9};
    }
  for (int len = (3); len <= (n); ++len) {
    for (int i = (1); i <= (n); ++i) {
      int j = i + len - 1;
      if (j > n) break;
      for (int k = (i + 1); k <= (j - 1); ++k) {
        node cur = (node){
            f[i][k].area + f[k][j].area +
                cross(x[k] - x[i], y[k] - y[i], x[j] - x[i], y[j] - y[i]),
            f[i][k].cnt + f[k][j].cnt};
        if (cur.area >= mid) {
          cur.area = 0, ++cur.cnt;
        }
        f[i][j] = max(f[i][j], cur);
      }
    }
  }
  return f[1][n].cnt >= K;
}
int main() {
  scanf("%d%d", &n, &K), ++K;
  for (int i = (1); i <= (n); ++i) scanf("%d%d", &x[i], &y[i]);
  LL l = 0, r = 1e18, ans = 0;
  chk(11);
  while (l <= r) {
    LL mid = (l + r) >> 1;
    if (chk(mid))
      ans = mid, l = mid + 1;
    else
      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;
using LL = long long;
const int N = 205;
int n, K, x[N], y[N];
struct node {
  LL area;
  int cnt;
  bool operator<(const node& k) const {
    return cnt != k.cnt ? cnt < k.cnt : area < k.area;
  }
} f[N][N];
LL cross(LL a, LL b, LL c, LL d) { return a * d - b * c; }
bool chk(LL mid) {
  for (int i = (1); i <= (n); ++i)
    for (int j = (i + 2); j <= (n); ++j) {
      f[i][j] = (node){(LL)1e18, (int)-1e9};
    }
  for (int len = (3); len <= (n); ++len) {
    for (int i = (1); i <= (n); ++i) {
      int j = i + len - 1;
      if (j > n) break;
      for (int k = (i + 1); k <= (j - 1); ++k) {
        node cur = (node){
            f[i][k].area + f[k][j].area +
                cross(x[k] - x[i], y[k] - y[i], x[j] - x[i], y[j] - y[i]),
            f[i][k].cnt + f[k][j].cnt};
        if (cur.area >= mid) {
          cur.area = 0, ++cur.cnt;
        }
        f[i][j] = max(f[i][j], cur);
      }
    }
  }
  return f[1][n].cnt >= K;
}
int main() {
  scanf("%d%d", &n, &K), ++K;
  for (int i = (1); i <= (n); ++i) scanf("%d%d", &x[i], &y[i]);
  LL l = 0, r = 1e18, ans = 0;
  chk(11);
  while (l <= r) {
    LL mid = (l + r) >> 1;
    if (chk(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```