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;
	vector<long long> x(n),y(n);
	vector<pair<int,int> >xy(n);
	for(int i=0;i<n;++i) {
		cin >> x[i] >> y[i];
		xy[i]=make_pair(x[i],y[i]);
	}
	sort(x.begin(),x.end());
	sort(y.begin(),y.end());
	long long ans=9e18;
	for(int x1=0;x1<n;++x1) {
		for(int x2=x1+1;x2<n;++x2) {
			for(int y1=0;y1<n;++y1) {
				for(int y2=y1+1;y2<n;++y2) {
					int cnt=0;
					for(int i=0;i<n;++i) {
						if(x[x1]<=xy[i].first&&xy[i].first<=x[x2]&&
						   y[y1]<=xy[i].second&&xy[i].second<=y[y2]) ++cnt;
					}
					if(cnt>=k) ans=min(ans,(x[x2]-x[x1])*(y[y2]-y[y1]));
				}
			}
		}
	}
	cout << ans << endl;
	
	return 0;
}