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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int N,K;
long long int x[51],y[51],x2[51],y2[51];
long long int ans;
int main(){
  cin >> N >> K;
  for(int i=0;i<N;i++){
    cin >> x[i] >> y[i];
    x2[i]=x[i];
    y2[i]=y[i];
  }
  sort(x,x+N);
  sort(y,y+N);
  ans=1LL*(x[N-1]-x[0])*(y[N-1]-y[0]);
  for(int xi=0;xi<N;xi++){
    for(int xj=xi+1;xj<N;xj++){
      for(int yi=0;yi<N;yi++){
	for(int yj=yi+1;yj<N;yj++){
	  
	  long long int lx=x[xi],rx=x[xj];
	  long long int by=y[yi],uy=y[yj];
	  int c=0;
	  for(int p=0;p<N;p++){
	    if(lx<=x2[p] && x2[p]<=rx && by<=y2[p] && y2[p]<=uy)c++;
	  }
	
	  if(c>=K)ans=min(ans,1LL*(rx-lx)*(uy-by));
	  
	}
      }
    }
  }
  cout << ans << endl;
  return 0;
}