#include <bits/stdc++.h>
using namespace std;
const int N = 205;
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) { x = _x, y = _y; }
  point operator-(const point& b) const { return point(x - b.x, y - b.y); }
};
pair<long long, long long> add(pair<long long, long long> a,
                               pair<long long, long long> b) {
  return {a.first + b.first, a.second + b.second};
}
long long cross(point a, point b) { return 1ll * a.x * b.y - 1ll * a.y * b.x; }
point a[N];
pair<long long, long long> f[N][N];
int n, K;
bool check(long long x) {
  for (int len = 3; len <= n; ++len)
    for (int l = 1; l + len - 1 <= n; ++l) {
      int r = l + len - 1;
      f[l][r] = {-1e9, 0};
      for (int k = l + 1; k < r; ++k) {
        pair<long long, long long> c = add(f[l][k], f[k][r]);
        c.second += cross(a[k] - a[l], a[r] - a[l]);
        if (c.second >= x) c.second = 0, c.first++;
        f[l][r] = max(f[l][r], c);
      }
    }
  return f[1][n].first >= K;
}
signed main() {
  cin >> n >> K;
  ++K;
  for (int i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y;
  long long l = 0, r = 1e18;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  cout << l;
}

### 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 int N = 205;
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) { x = _x, y = _y; }
  point operator-(const point& b) const { return point(x - b.x, y - b.y); }
};
pair<long long, long long> add(pair<long long, long long> a,
                               pair<long long, long long> b) {
  return {a.first + b.first, a.second + b.second};
}
long long cross(point a, point b) { return 1ll * a.x * b.y - 1ll * a.y * b.x; }
point a[N];
pair<long long, long long> f[N][N];
int n, K;
bool check(long long x) {
  for (int len = 3; len <= n; ++len)
    for (int l = 1; l + len - 1 <= n; ++l) {
      int r = l + len - 1;
      f[l][r] = {-1e9, 0};
      for (int k = l + 1; k < r; ++k) {
        pair<long long, long long> c = add(f[l][k], f[k][r]);
        c.second += cross(a[k] - a[l], a[r] - a[l]);
        if (c.second >= x) c.second = 0, c.first++;
        f[l][r] = max(f[l][r], c);
      }
    }
  return f[1][n].first >= K;
}
signed main() {
  cin >> n >> K;
  ++K;
  for (int i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y;
  long long l = 0, r = 1e18;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  cout << l;
}

```