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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		
		while(input.hasNext()){
			int N = input.nextInt();
			int K = input.nextInt();
			Point[] p = new Point[N];
			
			for(int i = 0; i < p.length; ++i){
				p[i] = new Point(input.nextInt(), input.nextInt());
			}
			
			long minRectangle = Long.MAX_VALUE;
			for(int x1 = 0; x1 < p.length; ++x1){
				for(int x2 = x1 + 1; x2 < p.length; ++x2){
					for(int y1 = 0; y1 < p.length; ++y1){
						for(int y2 = y1 + 1; y2 < p.length; ++y2){
							long MAX_X = Math.max(p[x1].x, p[x2].x);
							long MIN_X = Math.min(p[x1].x, p[x2].x);
							long MAX_Y = Math.max(p[y1].y, p[y2].y);
							long MIN_Y = Math.min(p[y1].y, p[y2].y);
							
							int countInside = 0;
							for(int i = 0; i < p.length; ++i){
								if(MIN_X <= p[i].x && p[i].x <= MAX_X && MIN_Y <= p[i].y && p[i].y <= MAX_Y){
									++countInside;
								}
								
								if(countInside > K){
									break;
								}
							}

							if(countInside == K){
								minRectangle = Math.min(minRectangle, (MAX_X - MIN_X) * (MAX_Y - MIN_Y));
							}
						}
					}
				}
			}
			
			System.out.println(minRectangle);
		}
	}
	
	private static class Point {
		int x;
		int y;
		
		Point (int x, int y){
			this.x = x;
			this.y = y;
		}
	}

}