#include <bits/stdc++.h>
using namespace std;
const int N = 204;
const long long INF = 1000000000000000000;
int n, k;
struct data {
  int ans;
  long long S;
  void chkmax(data x) {
    if (x.ans > ans || (x.ans == ans && x.S > S)) ans = x.ans, S = x.S;
  }
} f[N][N];
struct point {
  long long x, y;
} p[N];
point operator-(point a, point b) { return (point){a.x - b.x, a.y - b.y}; }
long long operator*(point a, point b) { return a.x * b.y - a.y * b.x; }
data operator+(data x, data y) { return (data){x.ans + y.ans, x.S + y.S}; }
bool check(long long S) {
  for (int j = 3; j <= n; ++j)
    for (int i = j - 2; i >= 1; --i) {
      f[i][j] = (data){0, 0};
      for (int k = i; k <= j; ++k) {
        data tmp = f[i][k] + f[k][j];
        tmp.S += abs((p[i] - p[k]) * (p[j] - p[k]));
        if (tmp.S >= S) tmp.S = 0, ++tmp.ans;
        f[i][j].chkmax(tmp);
      }
    }
  return f[1][n].ans > k;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%lld%lld", &p[i].x, &p[i].y);
  long long l = 1, r = INF;
  while (l <= r)
    check((l + r >> 1)) ? l = (l + r >> 1) + 1 : r = (l + r >> 1) - 1;
  printf("%lld\n", l - 1);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 204;
const long long INF = 1000000000000000000;
int n, k;
struct data {
  int ans;
  long long S;
  void chkmax(data x) {
    if (x.ans > ans || (x.ans == ans && x.S > S)) ans = x.ans, S = x.S;
  }
} f[N][N];
struct point {
  long long x, y;
} p[N];
point operator-(point a, point b) { return (point){a.x - b.x, a.y - b.y}; }
long long operator*(point a, point b) { return a.x * b.y - a.y * b.x; }
data operator+(data x, data y) { return (data){x.ans + y.ans, x.S + y.S}; }
bool check(long long S) {
  for (int j = 3; j <= n; ++j)
    for (int i = j - 2; i >= 1; --i) {
      f[i][j] = (data){0, 0};
      for (int k = i; k <= j; ++k) {
        data tmp = f[i][k] + f[k][j];
        tmp.S += abs((p[i] - p[k]) * (p[j] - p[k]));
        if (tmp.S >= S) tmp.S = 0, ++tmp.ans;
        f[i][j].chkmax(tmp);
      }
    }
  return f[1][n].ans > k;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%lld%lld", &p[i].x, &p[i].y);
  long long l = 1, r = INF;
  while (l <= r)
    check((l + r >> 1)) ? l = (l + r >> 1) + 1 : r = (l + r >> 1) - 1;
  printf("%lld\n", l - 1);
  return 0;
}

```