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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	class P implements Comparable<P>{
		long x;
		long y;
		P(long x, long y){
			this.x=x;
			this.y=y;
		}
		@Override
		public int compareTo(P arg0) {
			return Long.compare(this.x, arg0.x);
		}
	}
	public void run() {
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int k=sc.nextInt();
		ArrayList<P> ps = new ArrayList<>();
		for(int i=0; i<n; i++) {
			long x = sc.nextLong();
			long y = sc.nextLong();
			ps.add(new P(x,y));
		}

		Collections.sort(ps);
		long ans = Long.MAX_VALUE;
		for(int i=0; i<n-k+1; i++) {
			long minx=Long.MAX_VALUE;
			long maxx=-Long.MAX_VALUE;
			long miny=Long.MAX_VALUE;
			long maxy=-Long.MAX_VALUE;
			for(int j=i; j<i+k; j++) {
				P p = ps.get(j);
				minx=Math.min(minx, p.x);
				maxx=Math.max(maxx, p.x);
				miny=Math.min(miny, p.y);
				maxy=Math.max(maxy, p.y);
			}
			ans = Math.min(ans, (maxx-minx)*(maxy-miny));
		}
		System.out.println(ans);

		sc.close();
	}



}