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.io.IOException;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Created on 2017/10/20 7:35.
 */
public class Main {
    public static void main(String[] args) throws IOException {
        // BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        Scanner sc = new Scanner(System.in);

        while (sc.hasNextInt()) {
            int N = sc.nextInt();
            int K = sc.nextInt();
            int[] x = new int[N];
            int[] y = new int[N];

            for (int i = 0; i < N; i++) {
                x[i] = sc.nextInt();
                y[i] = sc.nextInt();
            }
            Arrays.sort(x);
            Arrays.sort(y);

            long ans = Long.MAX_VALUE;

            for (int yi = 0; yi < N; yi++) {
                int yFrom = y[yi];
                for (int yj = yi; yj < N; yj++) {
                    int yTo = y[yj];
                    int xi = 0;
                    int xj = 0;

                    while (xj < N) {
                        int xFrom = x[xi];
                        int xTo = x[xj];
                        int cnt = 0;
                        for (int i = 0; i < N; i++) {
                            if (xFrom <= x[i] && x[i] <= xTo && yFrom <= y[i] && y[i] <= yTo) {
                                cnt++;
                            }
                        }
                        if (cnt >= K) {
                            ans = Math.min(ans, (long) (xTo - xFrom) * (yTo - yFrom));
                            xi++;
                            xj = Math.max(xj, xi);
                        } else {
                            xj++;
                        }
                    }
                }
            }
            System.out.println(ans);
        }
    }
}