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>
using namespace std;

typedef long long ll;
int n,k;
ll x[55],y[55];

struct node{
	int x,y;
}arr[55];

int main(){
	memset(x,0,sizeof(x));
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++){
		ll u,v;
		scanf("%lld%lld",&u,&v);
		x[i]=u;	y[i]=v;
		arr[i].x=u;	arr[i].y=v;
	}
	sort(x,x+n);	sort(y,y+n);
	ll ans=5e18;
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			for(int L=0;L<n;L++){
				for(int R=L+1;R<n;R++){
					ll x1=x[i],x2=x[j];
					ll y1=y[L],y2=y[R];
					int tmp=0;
					for(int p=0;p<n;p++){
						if(arr[p].x>=x1&&arr[p].x<=x2&&arr[p].y>=y1&&arr[p].y<=y2){
							tmp++;
							if(tmp==k){
								ans=min(ans,(y2-y1)*(x2-x1));
								break;
							}
						}
					}
				}
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}