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.*;

class Main{
	static Scanner s=new Scanner(System.in);
	public static void main(String[] $){
		int n=s.nextInt(),k=s.nextInt();
		int[]x=new int[n],y=new int[n];
		for(int i=0;i<n;++i) {
			x[i]=s.nextInt();
			y[i]=s.nextInt();
		}
		long res=Long.MAX_VALUE;
		for(int i=0;i<n;++i) {
			for(int j=i+1;j<n;++j) {
				long l=Math.min(x[i],x[j]);
				long r=Math.max(x[i],x[j]);
				
				for(int p=0;p<n;++p) {
					for(int q=p+1;q<n;++q) {
						long t=Math.min(y[p],y[q]);
						long b=Math.max(y[p],y[q]);
						
						int c=0;
						for(int o=0;o<n;++o) {
							if(l<=x[o]&&x[o]<=r&&t<=y[o]&&y[o]<=b){
								++c;
							}
						}
						if(c>=k)
							res=Math.min(res,(r-l)*(b-t));
					}
				}
			}
		}
		System.out.println(res);
	}
}