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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int k = Integer.parseInt(sa[1]);
		long[] x = new long[n];
		long[] y = new long[n];
		for (int i = 0; i < n; i++) {
			sa = br.readLine().split(" ");
			x[i] = Integer.parseInt(sa[0]);
			y[i] = Integer.parseInt(sa[1]);
		}
		br.close();

		long ans = Long.MAX_VALUE;
		for (int a = 0; a < n; a++) {
			for (int b = 0; b < n; b++) {
				for (int c = 0; c < n; c++) {
					for (int d = 0; d < n; d++) {
						int cnt = 0;
						for (int i = 0; i < n; i++) {
							if (x[a] <= x[i] && x[i] <= x[b]
									&& y[c] <= y[i] && y[i] <= y[d]) {
								cnt++;
							}
						}
						if (cnt >= k) {
							ans = Math.min(ans, Math.abs((x[b] - x[a]) * (y[d] - y[c])));
						}
					}
				}
			}
		}
		System.out.println(ans);
	}
}