Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

long long min(long long x,long long y){
	if(x<y)return x;
	return y;
}

int compare(const void *a, const void *b){
	long long int *A = (long long int *)a;
	long long int *B = (long long int *)b;
	if (*A < *B) return 1;
	if (*A > *B) return -1;
	return 0;
}

int main(){
	long long N,K,i,j,k,l,m;
	scanf("%lld%lld",&N,&K);
	long long x[N],y[N],xsort[N],ysort[N];
	for(i=0;i<N;i++){
		scanf("%lld%lld",&x[i],&y[i]);
		xsort[i]=x[i];
		ysort[i]=y[i];
	}
	qsort(xsort,N,sizeof(long long),compare);
	qsort(ysort,N,sizeof(long long),compare);
	
	long long ans=(xsort[0]-xsort[N-1])*(ysort[0]-ysort[N-1]);
	
	for(i=0;i+K-1<N;i++){
		for(j=i+K-1;j<N;j++){
			for(k=0;k+K-1<N;k++){
				for(l=k+K-1;l<N;l++){
					long long c=0;
					for(m=0;m<N;m++)if(x[m]<=xsort[i] && x[m]>=xsort[j] && y[m]<=ysort[k] && y[m]>=ysort[l])c++;
					if(c>=K){
						ans=min(ans,(xsort[i]-xsort[j])*(ysort[k]-ysort[l]));
						break;
					}
				}
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}