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++14 (GCC 5.4.1) (judge verdict: Accepted)

//23:24
#include <stdio.h>

int x[50];
int y[50];
int min(int x, int y){ return (x < y)?x : y;}
int max(int x, int y){ return (x > y)?x : y;}
int main(){
	int N, K;
	scanf("%d %d",&N,&K);
	for(int i = 0; i < N; i++) scanf("%d %d",x+i,y+i);

	long long space = 4*1000000000LL*1000000000LL;
	for(int x1 = 0; x1 < N-1; x1++){
		for(int x2 = x1; x2 < N; x2++){
			for(int y1 = 0; y1 < N-1; y1++){
				for(int y2 = y1; y2 < N; y2++){
					long long  xs = min(x[x1],x[x2]);
					long long  xl = max(x[x1],x[x2]);
					long long  ys = min(y[y1],y[y2]);
					long long  yl = max(y[y1],y[y2]);
					int cnt=0;
					for(int n = 0; n < N; n++){
						if(xs <= x[n] && x[n] <= xl 
						&& ys <= y[n] && y[n] <= yl){ cnt++; }
					}				
					if(cnt >= K){
						if(space > (xl-xs)*(yl-ys)) space = (xl-xs)*(yl-ys);
					}
				}
			}
		}
	}
	printf("%lld\n",space);
}