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<cstring>
#include<algorithm>
const int N=58;
int x[N],y[N];
struct node
{
	int x,y;
}e[N];
inline long long min(long long a,long long b)
{
	return a<b?a:b;
}
int main()
{
	int n,k;
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		e[i].x=u,e[i].y=v;
		x[i]=u,y[i]=v;	
	}
	std::sort(x+1,x+1+n);
	std::sort(y+1,y+1+n);
	long long ans=(long long)(x[n]-x[1])*(y[n]-y[1]);
	for(int i=1;i<=n;i++)
	for(int j=i+1;j<=n;j++)
	for(int p=1;p<=n;p++)
	for(int q=p+1;q<=n;q++)
	{
		int sum=0;
		for(int k=1;k<=n;k++)
		if(e[k].x<=x[j]&&e[k].x>=x[i]&&e[k].y>=y[p]&&e[k].y<=y[q])	sum++;
		if(sum>=k)	ans=min(ans,(long long )(x[j]-x[i])*(y[q]-y[p]));	
	}
	printf("%lld\n",ans);
	return 0;
}