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;
vector<long>X,Y,Z;
long N,K,x[50],y[50],ans=LONG_MAX,i,j,k,L;
int main(){
  cin>>N>>K;
  for(i=0;i<N;i++){cin>>x[i]>>y[i];X.push_back(x[i]);Y.push_back(y[i]);}
  sort(X.begin(),X.end());sort(Y.begin(),Y.end());
  for(i=0;i<N;i++){
    for(j=i+1;j<N;j++){
      Z.clear();
      for(k=0;k<N;k++)if(X[i]<=x[k]&&x[k]<=X[j])Z.push_back(y[k]);
      sort(Z.begin(),Z.end());
      L=Z.size();
      for(k=0;k<=L-K;k++)ans=min(ans,(X[j]-X[i])*(Z[k+K-1]-Z[k]));
    }
  }
  cout<<ans<<endl;
}