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<cstdio>
#include<algorithm>
using namespace std;

#define ran 111

int n, m;
pair<int, int> v[ran], x[ran];

int cmp(pair<int, int> &a, pair<int, int> &b) {
	if (a.second != b.second) return a.second < b.second;
	return a.first < b.first;
}
void upd(long long &a, long long b) {
	if (a==-1 || a>b) a = b;
}

int main() {
	scanf("%d%d", &n, &m);
	for(int i=0;i<n;i++) {
		scanf("%d%d", &v[i].first, &v[i].second);
	}
	sort(v, v+n);
	
	long long ans = -1LL;
	for(int i=0;i+m-1<n;i++)
		for(int j=i+m-1;j<n;j++) {
			for(int k=i;k<=j;k++)
				x[k-i] = v[k];
			int len = j-i+1;
			sort(x, x+len, cmp);
			
			for(int l=0, r=m-1; r<len; l++, r++) 
				upd(ans, 1LL*(v[j].first - v[i].first) * (x[r].second - x[l].second));
		}
	printf("%lld\n", ans);
	
	return 0;
}