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: Wrong Answer)

import java.util.*;

public class Main {
	static long[] arrX;
	static long[] arrY;
	static int n;
	static int k;

	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		k = sc.nextInt();
		arrX = new long[n];
		arrY = new long[n];
		for (int i = 0; i < n; i++) {
			arrX[i] = sc.nextLong();
			arrY[i] = sc.nextLong();
		}
		long min = Long.MAX_VALUE;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				long sqr = Math.abs(arrX[i] - arrX[j]) * Math.abs(arrY[i] - arrY[j]);
              System.out.println(sqr);
				if (sqr > min) {
					continue;
				}
				if (isOver(Math.min(arrX[i], arrX[j]), Math.max(arrX[i], arrX[j]), Math.min(arrY[i], arrY[j]), Math.max(arrY[i], arrY[j]))) {
					min = sqr;
				}
			}
		}
		System.out.println(min);
	}
	
	static boolean isOver(long x1, long x2, long y1, long y2) {
		int count = 0;
		for (int i = 0; i < n; i++) {
			if (arrX[i] >= x1 && arrX[i] <= x2 && arrY[i] >= y1 && arrY[i] <= y2) {
				count++;
			}
		}
		return (count >= k);
	}
}