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>
#define For(i,n) for(int i=0; i<n; i++)
using namespace std;

int main(){
  int N,K;
  cin >> N >> K;
  long long x[N],y[N],xx[N],yy[N];
  long long mnX,mnY,mxX,mxY;
  long long size;
  For(i,N){
    cin >> x[i] >> y[i];
    xx[i] = x[i],yy[i] = y[i];
  }
  sort(x,x+N);
  sort(y,y+N);
  size = (x[N-1]-x[0])*(y[N-1]-y[0]);
  For(i,N-1) for(int j=i+1; j<N; j++){
    For(k,N-1) for(int l=k+1; l<N; l++){
      int num=0;
      mnX = x[i], mxX = x[j];
      mnY = y[k], mxY = y[l];
      for(int p=0; p<N; p++){
        if(mnX<=xx[p] && xx[p]<=mxX && mnY<=yy[p] && yy[p]<=mxY){
          num++;
        }
        if(num>=K) size = min(size, (mxX-mnX)*(mxY-mnY));
      }
    }
  }
  cout << size << endl;
  return 0;
}