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 <limits.h>

int main(void) {
	int i, ix, jx, iy, jy, l, m, N, K;
	long long x[50], y[50], lx, ly, sx, sy, R, S;
	scanf("%d%d", &N, &K);
	for (i=0;i<N;i++) {
		scanf("%lld%lld", &x[i], &y[i]);
	}

	S = LLONG_MAX;
	for (ix=1;ix<N;ix++) {
		for (jx=0;jx<ix;jx++) {
			if (x[ix] < x[jx]) {
				lx = x[jx]; sx = x[ix];
			} else {
				lx = x[ix]; sx = x[jx];				
			}
			for (iy=1;iy<N;iy++) {
				for (jy=0;jy<N;jy++) {
					if (y[iy] < y[jy]) {
						ly = y[jy]; sy = y[iy];
					} else {
						ly = y[iy]; sy = y[jy];				
					}
					R = (lx-sx)*(ly-sy);
					m = 0;
					for (l=0;l<N;l++) {
						if (x[l] <= lx && x[l] >= sx && y[l] <= ly && y[l] >= sy) m++;
					}
					if (m>=K) {
						if (R<S) S = R;
					}							
				}
			}
		}
	}

	printf("%lld\n", S);
	return 0;
}