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 n, K;
vector<int> x, y, v, w;
long long ans = 9e18;

int main(){
//	cin.tie(0);
//	ios::sync_with_stdio(false);
	cin >> n >> K;
	for(int i=0; i<n; ++i){
		int s, t;
		cin >> s >> t;
		x.push_back(s);
		y.push_back(t);
	}
	v = x;
	w = y;
	sort(v.begin(), v.end());
	sort(w.begin(), w.end());
	for(int i=0; i<n; ++i)
		for(int j=i+1; j<n; ++j)
			for(int k=0; k<n; ++k)
				for(int l=k+1; l<n; ++l){
					int tmp = 0;
					for(int m=0; m<n; ++m)
						if(v[i] <= x[m] && x[m] <= v[j] && w[k] <= y[m] && y[m] <= w[l])
							++tmp;
					if(tmp >= K)
						ans = min(ans, (long long)(v[j] - v[i]) * (w[l] - w[k]));
				}
	cout << ans << "\n";
}