#include <bits/stdc++.h>
using int64 = long long;
const int N = 205;
int n, k;
int next[N];
int64 S;
int64 x[N], y[N];
std::pair<int, int64> dp[N][N];
int check(int64 s) {
  S = s;
  int d = 0;
  for (int dis = 1; dis < n; ++dis) {
    for (int r = 0; r < n; ++r) {
      int l = (r + dis) % n;
      if (dis == 1 && (dp[l][r] = std::make_pair(0, 0), 1)) continue;
      dp[l][r] = std::make_pair(-1, -1);
      for (int k = next[r]; k != l; k = next[k]) {
        int64 s = (x[r] - x[l]) * (y[k] - y[l]) - (x[k] - x[l]) * (y[r] - y[l]);
        auto s1 = dp[l][k], s2 = dp[k][r];
        int p1 = s1.first + s2.first;
        int64 p2 = s1.second + s2.second + s;
        if (p2 >= S) p2 = 0, ++p1;
        dp[l][r] = std::max(dp[l][r], std::make_pair(p1, p2));
      }
    }
  }
  for (int i = 0; i < n; ++i) d = std::max(d, dp[i][next[i]].first);
  return d;
}
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
  std::cin >> n >> k;
  for (int i = 0; i < n; ++i) std::cin >> x[i] >> y[i], next[i] = (i + 1) % n;
  int64 l = 0, r = (1ll << 60), mid;
  while (l + 1 < r) (check(l + r >> 1) >= k + 1 ? l : r) = l + r >> 1;
  std::cout << l << std::endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 int64 = long long;
const int N = 205;
int n, k;
int next[N];
int64 S;
int64 x[N], y[N];
std::pair<int, int64> dp[N][N];
int check(int64 s) {
  S = s;
  int d = 0;
  for (int dis = 1; dis < n; ++dis) {
    for (int r = 0; r < n; ++r) {
      int l = (r + dis) % n;
      if (dis == 1 && (dp[l][r] = std::make_pair(0, 0), 1)) continue;
      dp[l][r] = std::make_pair(-1, -1);
      for (int k = next[r]; k != l; k = next[k]) {
        int64 s = (x[r] - x[l]) * (y[k] - y[l]) - (x[k] - x[l]) * (y[r] - y[l]);
        auto s1 = dp[l][k], s2 = dp[k][r];
        int p1 = s1.first + s2.first;
        int64 p2 = s1.second + s2.second + s;
        if (p2 >= S) p2 = 0, ++p1;
        dp[l][r] = std::max(dp[l][r], std::make_pair(p1, p2));
      }
    }
  }
  for (int i = 0; i < n; ++i) d = std::max(d, dp[i][next[i]].first);
  return d;
}
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
  std::cin >> n >> k;
  for (int i = 0; i < n; ++i) std::cin >> x[i] >> y[i], next[i] = (i + 1) % n;
  int64 l = 0, r = (1ll << 60), mid;
  while (l + 1 < r) (check(l + r >> 1) >= k + 1 ? l : r) = l + r >> 1;
  std::cout << l << std::endl;
  return 0;
}

```