Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999


Output

3999999996000000001

Solution (in JAVA):

Solution:
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();
		
		int[] a = new int[4];
		long[] x = new long[n];
		long[] y = new long[n];
		
		for(int i = 0; i < n; i++)
		{
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
		}
		
		int[] b = new int[4];
		long best = 8999999999999999999l;
		
		for(a[0] = 0; a[0] < n; a[0]++)
			for(a[1] = a[0]; a[1] < n; a[1]++)
				for(a[2] = a[1]; a[2] < n; a[2]++)
					for(a[3] = a[2]; a[3] < n; a[3]++)
					{
						long y0 = 2000000000;
						long y1 = -2000000000;
						long x0 = 2000000000;
						long x1 = -2000000000;
						for(int i = 0; i < 4; i++)
						{
							x0=Math.min(x0, x[a[i]]);
							x1=Math.max(x1, x[a[i]]);
							y0=Math.min(y0, y[a[i]]);
							y1=Math.max(y1, y[a[i]]);
						}
						int c = 0;
						for(int i = 0; i < n; i++)
						{
							if(x[i]>=x0&&x[i]<=x1&&y[i]>=y0&&y[i]<=y1)
								c++;
						}
						long area = (x1-x0)*(y1-y0);
						if(c>=k&&area<best)
						{
							for(int i = 0; i < 4; i++)
								b[i] = a[i];
							best = area;
						}
					}
			
		System.out.println(best);
	}

}