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 <iostream>
#include <algorithm>
using namespace std;

using ll = long long;

int main()
{
  int n, k;
  cin >> n >> k;

  int x[50], y[50];
  int xx[50], yy[50];
  for(int i = 0; i < n; i++){
    cin >> x[i] >> y[i];
    xx[i] = x[i];
    yy[i] = y[i];
  }

  sort(xx, xx+n);
  sort(yy, yy+n);

  ll ans = 1LL*(xx[n-1]-xx[0])*(yy[n-1]-yy[0]);
  for(int l = 0; l < n; l++){
    for(int r = l+1; r < n; r++){
      for(int b = 0; b < n; b++){
        for(int t = b+1; t < n; t++){
          int num = 0;
          for(int i = 0; i < n; i++){
            if(xx[l] <= x[i] && x[i] <= xx[r] && yy[b] <= y[i] && y[i] <= yy[t]){
              num++;
            }
          }
          if(num >= k) ans = min(ans, 1LL*(xx[r]-xx[l])*(yy[t]-yy[b]));
        }
      }
    }
  }

  cout << ans << endl;

  return 0;
}