#include <bits/stdc++.h>
struct point {
  int x, y;
  point() : x(), y() {}
  point(int _x, int _y) : x(_x), y(_y) {}
  point operator-(const point &rhs) const {
    return point(x - rhs.x, y - rhs.y);
  }
  friend long long cross(const point &a, const point &b) {
    return 1ll * a.x * b.y - 1ll * a.y * b.x;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, k;
  std::cin >> n >> k;
  ++k;
  std::vector<point> a(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i].x >> a[i].y;
  }
  auto check = [&](long long x) {
    std::vector<std::vector<std::pair<int, long long>>> f(
        n, std::vector<std::pair<int, long long>>(n));
    for (int i = n - 1; i >= 0; --i) {
      for (int j = i + 2; j < n; ++j) {
        for (int k = i + 1; k < j; ++k) {
          long long s = cross(a[k] - a[i], a[j] - a[i]);
          s += f[i][k].second + f[k][j].second;
          f[i][j] = std::max(
              f[i][j],
              std::pair<int, long long>(
                  f[i][k].first + f[k][j].first + (s >= x), s >= x ? 0ll : s));
        }
      }
    }
    return f[0][n - 1].first >= k;
  };
  long long l = 1, r = 1e18;
  while (l + 1 < r) {
    long long mid = (l + r) >> 1;
    if (check(mid)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  std::cout << l << "\n";
}

### 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>
struct point {
  int x, y;
  point() : x(), y() {}
  point(int _x, int _y) : x(_x), y(_y) {}
  point operator-(const point &rhs) const {
    return point(x - rhs.x, y - rhs.y);
  }
  friend long long cross(const point &a, const point &b) {
    return 1ll * a.x * b.y - 1ll * a.y * b.x;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, k;
  std::cin >> n >> k;
  ++k;
  std::vector<point> a(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i].x >> a[i].y;
  }
  auto check = [&](long long x) {
    std::vector<std::vector<std::pair<int, long long>>> f(
        n, std::vector<std::pair<int, long long>>(n));
    for (int i = n - 1; i >= 0; --i) {
      for (int j = i + 2; j < n; ++j) {
        for (int k = i + 1; k < j; ++k) {
          long long s = cross(a[k] - a[i], a[j] - a[i]);
          s += f[i][k].second + f[k][j].second;
          f[i][j] = std::max(
              f[i][j],
              std::pair<int, long long>(
                  f[i][k].first + f[k][j].first + (s >= x), s >= x ? 0ll : s));
        }
      }
    }
    return f[0][n - 1].first >= k;
  };
  long long l = 1, r = 1e18;
  while (l + 1 < r) {
    long long mid = (l + r) >> 1;
    if (check(mid)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  std::cout << l << "\n";
}

```