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() {
	long long n,t,i,j,k,l,m;
	cin >> n >> t;
	vector<pair<long long, long long>> a(n);
	vector<long long> x(n),y(n);
	for(i=0; i<n; i++){
		cin >> a[i].first >> a[i].second;
		x[i]=a[i].first; y[i]=a[i].second;
	}
	sort(x.begin(),x.end());
	sort(y.begin(),y.end());
	
	long long ans=4000000000000000000,cur;
	for(i=0; i<n-1; i++){
		for(j=i+1; j<n; j++){
			for(k=0; k<n-1; k++){
				for(l=k+1; l<n; l++){
					cur=0;
					for(m=0; m<n; m++){
						if(a[m].first>=x[i] && a[m].first<=x[j] && a[m].second>=y[k] && a[m].second<=y[l]){
							cur++;
						}
					}
					if(cur>=t){
						ans=min(ans,(x[j]-x[i])*(y[l]-y[k]));
					}
				}
			}
			
		}
	}
	cout << ans;
	
}