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>
long long int max(long long int a, long long int b)
{
	if (a > b)
		return a;
	else
		return b;
}
long long int min(long long int a, long long int b)
{
	if (a < b)
		return a;
	else
		return b;
}
int main()
{
	int n, k;
	scanf("%d %d", &n, &k);
	int i;
	long long int x[55], y[55];
	for (i = 0; i < n; i++)
		scanf("%lld %lld", &x[i], &y[i]);
	int a, b, c, d;
	long long int minx, miny, maxx, maxy;
	int count;
	long long int ans = 1e18 * 4 + 1e9;
	for (a = 0; a < n; a++)
	{
		for (b = 0; b < n; b++)
		{
			for (c = 0; c < n; c++)
			{
				for (d = 0; d < n; d++)
				{
					minx = min(min(x[a], x[b]), min(x[c], x[d]));
					miny = min(min(y[a], y[b]), min(y[c], y[d]));
					maxx = max(max(x[a], x[b]), max(x[c], x[d]));
					maxy = max(max(y[a], y[b]), max(y[c], y[d]));
					count = 0;
					for (i = 0; i < n; i++)
						if (minx <= x[i] && x[i] <= maxx && miny <= y[i] && y[i] <= maxy)
							count++;
					if (count >= k)
						ans = min(ans, (maxx - minx) * (maxy - miny));
				}
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}