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;
int main(){
    int n,k;
    cin>>n>>k;
    long long x[n],y[n];
    vector<pair<long long,long long>>p(n);
    for(int i=0;i<n;i++){
        cin>>x[i]>>y[i];
        p[i]={x[i],y[i]};
    }
    sort(x,x+n);
    sort(y,y+n);
    long long m=5000000000000000000;
    for(int l=0;l<n-1;l++){
        for(int r=l+1;r<n;r++){
            for(int d=0;d<n-1;d++){
                for(int u=d+1;u<n;u++){
                    int point=0;
                    for(int i=0;i<n;i++)if(x[l]<=p[i].first && p[i].first<=x[r] && y[d]<=p[i].second && p[i].second<=y[u])point++;
                    if(point>=k)m=min(m,(x[r]-x[l])*(y[u]-y[d]));
                }
            }
        }
    }
    cout<<m<<endl;
    return 0;
}