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>
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define debug(x) cout<<#x<<"="<<x<<endl;
#define LL long long 
#define pll pair<LL ,LL >
#define fi first
#define se second
using namespace std;
const int MAX=55;
pll pt[MAX];
int xa[MAX];
int ya[MAX];
int main(void)
{
	int n,m;
	scanf("%d%d",&n,&m);
	rep(i,0,n)
	{
		scanf("%lld%lld",&pt[i].fi,&pt[i].se);
		xa[i]=pt[i].fi;
		ya[i]=pt[i].se;
	}
	sort(xa,xa+n);
	sort(ya,ya+n);
	LL mx=1e19; 
	rep(i,0,n)
	{
		rep(j,i,n)
		{
			rep(k,0,n)
			{
				rep(l,k,n)
				{
					int cnt=0;
					rep(tmp,0,n)
					{
						if(pt[tmp].fi>=xa[i]&&pt[tmp].fi<=xa[j]&&pt[tmp].se>=ya[k]&&pt[tmp].se<=ya[l])cnt++;
					}
					if(cnt>=m)mx=min(mx,(xa[j]-xa[i])*1ll*(ya[l]-ya[k]));
				}
			} 
		} 
	}
	printf("%lld",mx);
	return 0;
}