#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const unsigned gen_seed =
    std::chrono::system_clock::now().time_since_epoch().count();
std::mt19937_64 gen(gen_seed);
std::vector<int> x, y;
long long s32(long long x1, long long y1, long long x2, long long y2,
              long long x3, long long y3) {
  return (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
}
long long s3(int i, int j, int k) {
  return s32(x[i], y[i], x[j], y[j], x[k], y[k]);
}
bool cw(int i, int j, int k) { return s3(i, j, k) < 0; }
bool ccw(int i, int j, int k) { return s3(i, j, k) > 0; }
long long v3s[200][200][200];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  x.resize(n);
  y.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &x[i], &y[i]);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++) v3s[i][j][k] = s3(i, j, k);
  long long l = 0;
  long long r = 1e18;
  while (r - l > 1) {
    long long m = (r + l) / 2;
    std::vector<std::vector<std::pair<long long, long long> > > d(
        n, std::vector<std::pair<long long, long long> >(n, make_pair(0, 0)));
    for (int l = 2; l < n; l++) {
      for (int s = 0; s < n - l; s++) {
        int e = s + l;
        for (int f = s + 1; f < e; f++) {
          std::pair<long long, long long> cand =
              make_pair(d[s][f].first + d[f][e].first,
                        d[s][f].second + d[f][e].second + v3s[s][f][e]);
          if (cand.second >= m) {
            cand.first++;
            cand.second = 0;
          }
          d[s][e] = max(d[s][e], cand);
        }
      }
    }
    if (d[0][n - 1].first >= k + 1)
      l = m;
    else
      r = m;
  }
  cout << l;
}

### Prompt

Please create a solution in CPP 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;
const long long mod = 1000000007;
const unsigned gen_seed =
    std::chrono::system_clock::now().time_since_epoch().count();
std::mt19937_64 gen(gen_seed);
std::vector<int> x, y;
long long s32(long long x1, long long y1, long long x2, long long y2,
              long long x3, long long y3) {
  return (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
}
long long s3(int i, int j, int k) {
  return s32(x[i], y[i], x[j], y[j], x[k], y[k]);
}
bool cw(int i, int j, int k) { return s3(i, j, k) < 0; }
bool ccw(int i, int j, int k) { return s3(i, j, k) > 0; }
long long v3s[200][200][200];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  x.resize(n);
  y.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &x[i], &y[i]);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++) v3s[i][j][k] = s3(i, j, k);
  long long l = 0;
  long long r = 1e18;
  while (r - l > 1) {
    long long m = (r + l) / 2;
    std::vector<std::vector<std::pair<long long, long long> > > d(
        n, std::vector<std::pair<long long, long long> >(n, make_pair(0, 0)));
    for (int l = 2; l < n; l++) {
      for (int s = 0; s < n - l; s++) {
        int e = s + l;
        for (int f = s + 1; f < e; f++) {
          std::pair<long long, long long> cand =
              make_pair(d[s][f].first + d[f][e].first,
                        d[s][f].second + d[f][e].second + v3s[s][f][e]);
          if (cand.second >= m) {
            cand.first++;
            cand.second = 0;
          }
          d[s][e] = max(d[s][e], cand);
        }
      }
    }
    if (d[0][n - 1].first >= k + 1)
      l = m;
    else
      r = m;
  }
  cout << l;
}

```