Problem: We have N points in a two-dimensional plane.
The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i).
Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior.
Here, points on the sides of the rectangle are considered to be in the interior.
Find the minimum possible area of such a rectangle.

Constraints

* 2 \leq K \leq N \leq 50
* -10^9 \leq x_i,y_i \leq 10^9 (1 \leq i \leq N)
* x_i≠x_j (1 \leq i<j \leq N)
* y_i≠y_j (1 \leq i<j \leq N)
* All input values are integers. (Added at 21:50 JST)

Input

Input is given from Standard Input in the following format:


N K
x_1 y_1
:
x_{N} y_{N}


Output

Print the minimum possible area of a rectangle that satisfies the condition.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999


Output

3999999996000000001

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n, k;
  cin >> n >> k;
  
  vector<pair<ll, ll>> x(n);
  for(int i = 0; i < n; i++) cin >> x.at(i).first >> x.at(i).second;
  
  sort(x.begin(), x.end());
  
  ll ans = 4e18 + 1;
  for(int i = 0; i < n - 1; i++){
    for(int j = i + 1; j < n; j++){
      if(j - i + 1 < k) continue;
      
      vector<ll> y;
      for(int l = i; l <= j; l++) y.push_back(x.at(l).second);
      
      sort(y.begin(), y.end());
      
      for(int l = 0; l <= j - i + 1 - k; l++){
        ans = min(ans, (x.at(j).first - x.at(i).first) * (y.at(l + k - 1) - y.at(l)));
      }
    }
  }
  
  cout << ans << '\n';
}