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)

/* Coached by Dukkha */
#include <stdio.h>
#include <stdlib.h>

#define N 50
#define INF 0x7fffffff

long long min(long long a, long long b) { return a < b ? a : b; }

int xx[N], yy[N], ii[N], xx_[N], n, k;

int compare(const void *a, const void *b) {
	int i = *(int *) a;
	int j = *(int *) b;

	return xx[i] - xx[j];
}

int solve(int y1, int y2) {
	int i, cnt = 0, w = INF;

	for (i = 0; i < n; i++) {
		int x = xx[ii[i]], y = yy[ii[i]];

		if (y1 <= y && y <= y2)
			xx_[cnt++] = x;
		if (cnt >= k)
			w = min(w, xx_[cnt - 1] - xx_[cnt - k]);
	}
	return w;
}

int main() {
	int i, j;
	long long ans;

	scanf("%d%d", &n, &k);
	for (i = 0; i < n; i++)
		scanf("%d%d", &xx[i], &yy[i]);
	for (i = 0; i < n; i++)
		ii[i] = i;
	qsort(ii, n, sizeof *ii, compare);
	ans = (long long) INF * INF;
	for (i = 0; i < n; i++)
		for (j = i + 1; j < n; j++) {
			int yi = yy[i], yj = yy[j], w;
			long long a;

			if (yi > yj) {
				int tmp;
				
				tmp = yi, yi = yj, yj = tmp;
			}
			w = solve(yi, yj);
			if (w == INF)
				continue;
			a = (long long) w * (yj - yi);
			ans = min(ans, a);
		}
	printf("%lld\n", ans);
	return 0;
}