#include <bits/stdc++.h>
using namespace std;
using ll = long long;
struct pt {
  ll x, y;
  pt(ll x_ = 0, ll y_ = 0) : x(x_), y(y_) {}
  pt operator+(pt p) const { return pt(x + p.x, y + p.y); }
  pt operator-(pt p) const { return pt(x - p.x, y - p.y); }
  friend ll cross(pt a, pt b) { return a.x * b.y - a.y * b.x; }
  friend ll cross(pt a, pt b, pt o) { return cross(a - o, b - o); }
};
void solve() {
  int n, k;
  cin >> n >> k;
  vector<pt> p(n);
  for (auto &[x, y] : p) {
    cin >> x >> y;
  }
  vector<vector<ll>> area(n, vector<ll>(n));
  for (int i = 0; i < n; i++) {
    for (int d = 1; d <= n - 1; d++) {
      int j1 = (i + d - 1) % n, j2 = (i + d) % n;
      area[i][j2] = area[i][j1] + cross(p[i], p[j1], p[j2]);
    }
  }
  auto check = [&](ll t) -> bool {
    vector<vector<int>> dp(n, vector<int>(n, 0));
    vector<vector<int>> minl(n + 1, vector<int>(n, 1e9));
    for (int d = 2; d <= n - 1; d++) {
      for (int l = 0; l < n; l++) {
        int r = (l + d) % n;
        auto &cur = dp[l][r];
        const auto ar = area[l][r] - t;
        for (int dl = 1; dl < d; dl++) {
          int m = (l + dl) % n;
          ll arl = dp[l][m] == 0 ? 0 : area[l][m];
          ll arr = dp[m][r] == 0 ? 0 : area[m][r];
          cur = max(cur, dp[l][m] + dp[m][r] + (ar >= arl + arr));
          int le = minl[cur][m];
          if (le + dl < d && ar >= area[m][(m + le) % n]) cur++;
        }
        minl[cur][l] = min(minl[cur][l], d);
      }
    }
    for (int l = 0; l < n; l++) {
      for (int r = 0; r < n; r++) {
        if (dp[l][r] + dp[r][l] >= k + 1) return true;
      }
    }
    return false;
  };
  if (k == 0) {
    cout << area[0][n - 1] << '\n';
    return;
  }
  ll l = 0, r = 1LL << 62;
  while (l < r - 1) {
    ll m = l + (r - l) / 2;
    if (check(m)) {
      l = m;
    } else {
      r = m;
    }
  }
  cout << l << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int owo = 1;
  while (owo--) solve();
}

### Prompt

Your task is to create 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;
using ll = long long;
struct pt {
  ll x, y;
  pt(ll x_ = 0, ll y_ = 0) : x(x_), y(y_) {}
  pt operator+(pt p) const { return pt(x + p.x, y + p.y); }
  pt operator-(pt p) const { return pt(x - p.x, y - p.y); }
  friend ll cross(pt a, pt b) { return a.x * b.y - a.y * b.x; }
  friend ll cross(pt a, pt b, pt o) { return cross(a - o, b - o); }
};
void solve() {
  int n, k;
  cin >> n >> k;
  vector<pt> p(n);
  for (auto &[x, y] : p) {
    cin >> x >> y;
  }
  vector<vector<ll>> area(n, vector<ll>(n));
  for (int i = 0; i < n; i++) {
    for (int d = 1; d <= n - 1; d++) {
      int j1 = (i + d - 1) % n, j2 = (i + d) % n;
      area[i][j2] = area[i][j1] + cross(p[i], p[j1], p[j2]);
    }
  }
  auto check = [&](ll t) -> bool {
    vector<vector<int>> dp(n, vector<int>(n, 0));
    vector<vector<int>> minl(n + 1, vector<int>(n, 1e9));
    for (int d = 2; d <= n - 1; d++) {
      for (int l = 0; l < n; l++) {
        int r = (l + d) % n;
        auto &cur = dp[l][r];
        const auto ar = area[l][r] - t;
        for (int dl = 1; dl < d; dl++) {
          int m = (l + dl) % n;
          ll arl = dp[l][m] == 0 ? 0 : area[l][m];
          ll arr = dp[m][r] == 0 ? 0 : area[m][r];
          cur = max(cur, dp[l][m] + dp[m][r] + (ar >= arl + arr));
          int le = minl[cur][m];
          if (le + dl < d && ar >= area[m][(m + le) % n]) cur++;
        }
        minl[cur][l] = min(minl[cur][l], d);
      }
    }
    for (int l = 0; l < n; l++) {
      for (int r = 0; r < n; r++) {
        if (dp[l][r] + dp[r][l] >= k + 1) return true;
      }
    }
    return false;
  };
  if (k == 0) {
    cout << area[0][n - 1] << '\n';
    return;
  }
  ll l = 0, r = 1LL << 62;
  while (l < r - 1) {
    ll m = l + (r - l) / 2;
    if (check(m)) {
      l = m;
    } else {
      r = m;
    }
  }
  cout << l << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int owo = 1;
  while (owo--) solve();
}

```