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;
import java.util.Arrays;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n = sc.nextInt();
		int t = sc.nextInt();
		Point[] p=new Point[n];
		for(int i = 0; i<n; i++){
			long x = sc.nextLong();
			long y = sc.nextLong();
			p[i]=new Point(x,y);
		}
		long ans=Long.MAX_VALUE;
		for(int i = 0; i<n; i++){
			for(int j = 0; j<n; j++){
				for(int k = 0; k<n; k++){
					for(int l = 0; l<n; l++){
						int count=0;
						for(int u=0; u<n; u++){
							if(p[u].x<=p[i].x && p[u].x>=p[j].x && p[u].y<=p[k].y && p[u].y>=p[l].y){
								count++;
							}
						}
						if(count>=t){
							ans=Math.min(ans,(p[i].x-p[j].x)*(p[k].y-p[l].y));
						}
					}
				}
			}
		}
		System.out.println(ans);
	}
	static class Point{
		long x;
		long y;
		public Point(long x,long y){
			this.x=x;
			this.y=y;
		}
	}
}