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;

void solve(long long N, long long K, vector<long long> x, vector<long long> y){
	long long res = LLONG_MAX;
	for (int i=0; i<N; i++) {
		for (int j=0; j<N; j++) {
			if (x[i]>=x[j]) continue;
			for (int k=0; k<N; k++) {
				for (int l=0; l<N; l++) {
					if (y[k]>=y[l]) continue;
					int count = 0;
					for (int m=0; m<N; m++) {
						if (x[i] <= x[m] && x[m] <= x[j] && y[k] <= y[m] && y[m] <= y[l]) count++;
						if (count >= K) res = min(res, (x[j]-x[i])*(y[l]-y[k]));
					}
				}
			}
		}
	}
	cout << res << endl;
}

int main(){	
	long long N;
	long long K;
	scanf("%lld",&N);
	vector<long long> y(N-1+1);
	vector<long long> x(N-1+1);
	scanf("%lld",&K);
	for(int i = 0 ; i <= N-1 ; i++){
		scanf("%lld",&x[i]);
		scanf("%lld",&y[i]);
	}
	solve(N, K, x, y);
	return 0;
}