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<string.h>
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=55;
int n,m;
vector<long long> x,y;
long long ax[maxn],ay[maxn];
int main(){
	int i,j,k,l,t;
	long long lx,ly,rx,ry;
	scanf("%d%d",&n,&m);
	for(i=0;i<n;i++){
		scanf("%lld%lld",&ax[i],&ay[i]);
		x.push_back(ax[i]);
		y.push_back(ay[i]);
	}
	sort(x.begin(),x.end());
	sort(y.begin(),y.end());
	long long ans=1LL*(x[n-1]-x[0])*(y[n-1]-y[0]);
	for(i=0;i<n;i++){
		for(j=i+1;j<n;j++){
			for(k=0;k<n;k++){
				for(l=k+1;l<n;l++){
					lx=x[i],rx=x[j];
					ry=y[l],ly=y[k];
					int num=0;
					for(t=0;t<n;t++){
						if(ax[t]>=lx&&ax[t]<=rx&&ay[t]>=ly&&ay[t]<=ry)
							num++;
					}
					if(num>=m)ans=min(ans,1LL*(rx-lx)*(ry-ly));
				}
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}