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.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        long[] x = new long[n];
        long[] y = new long[n];
        for (int i = 0 ; i < n ; i++) {
            x[i] = sc.nextLong();
            y[i] = sc.nextLong();
        }
        //二辺の組み合わせを総当たり
        long ans = Long.MAX_VALUE;
        for (int hl = 0 ; hl < n ; hl++) {
            for (int hr = hl + 1 ; hr < n ; hr++) {
                for (int wl = 0 ; wl < n ; wl++) {
                    for (int wr = wl + 1 ; wr < n ; wr++) {
                        long l = Math.min(x[wl], x[wr]);
                        long r = Math.max(x[wl], x[wr]);
                        long d = Math.min(y[hl], y[hr]);
                        long u = Math.max(y[hl], y[hr]);
                        int count = 0;
                        for (int i = 0 ; i < n ; i++) {
                            if (x[i] >= l && x[i] <= r && y[i] >= d && y[i] <= u ) {
                                count++;
                            }
                        }
                        if (count == k) ans = Math.min(ans, (r - l) * (u - d) );
                    }
                }
            }
        }

        System.out.println(ans);

    }

}