#include <bits/stdc++.h>
using namespace std;
int n, m;
struct Point {
  int x, y;
  Point(int _x = 0, int _y = 0) { x = _x, y = _y; }
  friend Point operator-(const Point &a, const Point &b) {
    return Point(a.x - b.x, a.y - b.y);
  }
  friend long long operator%(const Point &a, const Point &b) {
    return 1LL * a.x * b.y - 1LL * a.y * b.x;
  }
} p[404];
long long Z;
bool check(long long z) {
  static pair<int, long long> dp[404][404];
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n * 2 - 2; ++i) {
    long long w = (p[i + 1] - p[i]) % (p[i + 2] - p[i]);
    dp[i][i + 2] = w >= z ? make_pair(1, 0LL) : make_pair(0, w);
    if (dp[i][i + 2].first >= m) return true;
  }
  for (int len = 4; len <= n + 1; ++len) {
    for (int l = 1; l + len - 1 <= 2 * n; ++l) {
      int r = l + len - 1;
      dp[l][r] = make_pair(0, 0LL);
      for (int k = l + 1; k < r; ++k) {
        auto t = make_pair(
            dp[l][k].first + dp[k][r].first,
            dp[l][k].second + dp[k][r].second + (p[k] - p[l]) % (p[r] - p[l]));
        if (t.second >= z) ++t.first, t.second = 0;
        if (t.first >= m) return true;
        dp[l][r] = max(dp[l][r], t);
      }
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  ++m;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i].x >> p[i].y;
  }
  for (int i = 1; i <= n; ++i) {
    p[i + n] = p[i];
  }
  long long l = 1, r = 1e17 + 233, ans = 0;
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid, l = mid + 1;
    } else {
      r = mid;
    }
  }
  cout << ans << '\n';
  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;
int n, m;
struct Point {
  int x, y;
  Point(int _x = 0, int _y = 0) { x = _x, y = _y; }
  friend Point operator-(const Point &a, const Point &b) {
    return Point(a.x - b.x, a.y - b.y);
  }
  friend long long operator%(const Point &a, const Point &b) {
    return 1LL * a.x * b.y - 1LL * a.y * b.x;
  }
} p[404];
long long Z;
bool check(long long z) {
  static pair<int, long long> dp[404][404];
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n * 2 - 2; ++i) {
    long long w = (p[i + 1] - p[i]) % (p[i + 2] - p[i]);
    dp[i][i + 2] = w >= z ? make_pair(1, 0LL) : make_pair(0, w);
    if (dp[i][i + 2].first >= m) return true;
  }
  for (int len = 4; len <= n + 1; ++len) {
    for (int l = 1; l + len - 1 <= 2 * n; ++l) {
      int r = l + len - 1;
      dp[l][r] = make_pair(0, 0LL);
      for (int k = l + 1; k < r; ++k) {
        auto t = make_pair(
            dp[l][k].first + dp[k][r].first,
            dp[l][k].second + dp[k][r].second + (p[k] - p[l]) % (p[r] - p[l]));
        if (t.second >= z) ++t.first, t.second = 0;
        if (t.first >= m) return true;
        dp[l][r] = max(dp[l][r], t);
      }
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  ++m;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i].x >> p[i].y;
  }
  for (int i = 1; i <= n; ++i) {
    p[i + n] = p[i];
  }
  long long l = 1, r = 1e17 + 233, ans = 0;
  while (l < r) {
    long long mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid, l = mid + 1;
    } else {
      r = mid;
    }
  }
  cout << ans << '\n';
  return 0;
}

```