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.Scanner;

public class Main {

	public static void main(String[] args) {
		//問題入力
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[][] array = new int[n][2];

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < 2; j++) {
				array[i][j] = sc.nextInt();
			}
		}
		long space = 1L << 62;
		; //暫定的な面積(あとで表示するのでこれはforのブロックの外で宣言)
			//長方形の各辺の座標をarrayから探す
		int ceil, floor, left, right; //各辺の位置
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				for (int k1 = 0; k1 < n; k1++) {
					for (int l = 0; l < n; l++) {
						int counter = 0; //長方形に含まれる点の数をカウント
						for (int c = 0; c < n; c++) {
							if (array[c][0] >= array[i][0] && array[c][0] <= array[j][0] && array[c][1] >= array[k1][1]
									&& array[c][1] <= array[l][1]) {
								counter++;
							}
						}
						if (counter >= k) {
							space = Math.min(space,
									Math.abs((array[i][0] - array[i][1]) * (array[k1][1] - array[l][1])));
						}
					}
				}
			}
		}
		System.out.println(space);
		sc.close();
	}
}