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>
typedef long long ll;
using namespace std;

const int maxn=55;
int n,K,m,x[maxn],y[maxn],a[maxn];
ll ans=5e18;
int main(){
	scanf("%d%d",&n,&K);
	for (int i=1;i<=n;i++) scanf("%d%d",&x[i],&y[i]);
	for (int i=1;i<=n;i++)
	 for (int j=1;j<=n;j++)
	  if (y[i]<=y[j]){
	 	m=0;
		for (int k=1;k<=n;k++) if (y[i]<=y[k]&&y[k]<=y[j]) a[++m]=x[k];
		sort(a+1,a+1+m);
		for (int l=1,r=1;l<=m&&r<=m;l++){
			while (r<=m&&r-l+1<K) r++;
			if (r>m) break;
			ans=min(ans,(ll)(a[r]-a[l])*(y[j]-y[i]));
		}
	  }
	printf("%lld",ans);
	return 0;
}