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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
typedef long long LL;

using namespace std;
struct point
{
	LL x;
	LL y;
}Point[60]; 

bool cmp(struct point a,struct point b)
{
	return a.x<b.x;
}
LL my_min(LL a,LL b)
{
	return a<b ? a:b; 
}
int main()
{
	LL min=7e18;
	int N,K;
	scanf("%d %d",&N,&K);
	for(int i=0;i<N;i++)
		scanf("%lld %lld",&Point[i].x,&Point[i].y);
	sort(Point,Point+N,cmp);
	for(int i=0;i+K-1<N;i++)
	{
		LL dd[60];
		int len=0;
		for(int j=0;j<K-1;j++) 
			dd[len++]=Point[i+j].y;
		for(int l=K-1;i+l<N;l++)
		{
			dd[len++]=Point[i+l].y;
			sort(dd,dd+len);
			for(int q=0;q+K-1<len;q++)
				min=my_min(min,(Point[i+l].x-Point[i].x)*(dd[q+K-1]-dd[q]));
		}
	}
	printf("%lld\n",min);
	return 0;
}