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<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll x[55],y[55];
ll xx[55],yy[55];
int n,kk;
ll ans=9223372036854775807;
inline ll check(int u,int v,int k,int l)
{
ll x1=min(x[u],x[v]),x2=max(x[u],x[v]);
ll y1=min(y[k],y[l]),y2=max(y[k],y[l]);
int tot=0;
for(int i=1;i<=n;i++)
  if(x1<=x[i]&&x[i]<=x2&&y1<=y[i]&&y[i]<=y2) tot++;
return tot;
}
int main()
{
	scanf("%d%d",&n,&kk);
	for(int i=1;i<=n;i++)
	  scanf("%lld%lld",&x[i],&y[i]);
	for(int i=1;i<=n;i++)
	  for(int j=i;j<=n;j++)
	    for(int k=1;k<=n;k++)
	      for(int l=k;l<=n;l++)
			if(check(i,j,k,l)>=kk) ans=min(ans,1ll*abs(x[i]-x[j])*abs(y[k]-y[l]));
	printf("%lld",ans);
	return 0;
}