#include <bits/stdc++.h>
using namespace std;
const long long INF = 1LL << 61LL;
struct point {
  int x, y;
  point(int _x = 0, int _y = 0) {
    x = _x;
    y = _y;
  }
  point operator-(const point z) const { return point(x - z.x, y - z.y); }
} a[202];
struct val {
  long long num;
  long long remain;
  val(long long _a = 0, long long _b = 0) {
    num = _a;
    remain = _b;
  }
  val operator+(const val x) const {
    return val(num + x.num, remain + x.remain);
  }
  bool operator<(const val x) const {
    if (x.num == num) return remain < x.remain;
    return num < x.num;
  }
} f[202][202], tmp;
int n, m;
long long cross(point a, point b) {
  return abs(1LL * a.x * b.y - 1LL * a.y * b.x);
}
bool chk(long long x) {
  int i, j, k, len;
  for (i = 1; i <= n; ++i) {
    for (j = i + 2; j <= n; ++j) f[i][j] = val(-(INF >> 1), -(INF >> 1));
  }
  for (len = 2; len < n; ++len) {
    for (i = 1; i + len <= n; ++i) {
      j = i + len;
      for (k = i + 1; k < j; ++k) {
        tmp = f[i][k] + f[k][j];
        tmp.remain += cross(a[k] - a[j], a[i] - a[j]);
        if (tmp.remain >= x) {
          ++tmp.num;
          tmp.remain = 0;
        }
        if (f[i][j] < tmp) f[i][j] = tmp;
      }
    }
  }
  return f[1][n].num >= m;
}
int main() {
  int i, j;
  long long l = 0, r = INF, mid, ans = -1;
  scanf("%d%d", &n, &m);
  ++m;
  for (i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
  while (l <= r) {
    mid = (l + r) >> 1;
    if (chk(mid)) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 long long INF = 1LL << 61LL;
struct point {
  int x, y;
  point(int _x = 0, int _y = 0) {
    x = _x;
    y = _y;
  }
  point operator-(const point z) const { return point(x - z.x, y - z.y); }
} a[202];
struct val {
  long long num;
  long long remain;
  val(long long _a = 0, long long _b = 0) {
    num = _a;
    remain = _b;
  }
  val operator+(const val x) const {
    return val(num + x.num, remain + x.remain);
  }
  bool operator<(const val x) const {
    if (x.num == num) return remain < x.remain;
    return num < x.num;
  }
} f[202][202], tmp;
int n, m;
long long cross(point a, point b) {
  return abs(1LL * a.x * b.y - 1LL * a.y * b.x);
}
bool chk(long long x) {
  int i, j, k, len;
  for (i = 1; i <= n; ++i) {
    for (j = i + 2; j <= n; ++j) f[i][j] = val(-(INF >> 1), -(INF >> 1));
  }
  for (len = 2; len < n; ++len) {
    for (i = 1; i + len <= n; ++i) {
      j = i + len;
      for (k = i + 1; k < j; ++k) {
        tmp = f[i][k] + f[k][j];
        tmp.remain += cross(a[k] - a[j], a[i] - a[j]);
        if (tmp.remain >= x) {
          ++tmp.num;
          tmp.remain = 0;
        }
        if (f[i][j] < tmp) f[i][j] = tmp;
      }
    }
  }
  return f[1][n].num >= m;
}
int main() {
  int i, j;
  long long l = 0, r = INF, mid, ans = -1;
  scanf("%d%d", &n, &m);
  ++m;
  for (i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
  while (l <= r) {
    mid = (l + r) >> 1;
    if (chk(mid)) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```