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>
#include <string>
#include <vector>
using namespace std;

int main(){
  int n,k;
  cin>>n>>k;
  vector<long long> x(n),y(n);
  vector<long long> xa,ya;

  for(int i=0;i<n;i++){
    cin>>x[i]>>y[i];
    xa.push_back(x[i]);
    ya.push_back(y[i]);
  }

  sort(xa.begin(),xa.end());
  sort(ya.begin(),ya.end());

  long long ans=(xa[n-1]-xa[0])*(ya[n-1]-ya[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++){
          int num=0;
          for(int i=0;i<n;i++){
            if(xa[xi]<=x[i]&&xa[xj]>=x[i]&&ya[yi]<=y[i]&&ya[yj]>=y[i]) num++;
          }
          if(num>=k){
            ans=min(ans,(xa[xj]-xa[xi])*(ya[yj]-ya[yi]));
          }
        }
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}