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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		List<int[]> list = new ArrayList<>();
		for(int i=0;i<n;i++) {
			int[] tmp = {in.nextInt(),in.nextInt()};
			list.add(tmp);
		}
		list.sort((x1,x2)->x1[0]-x2[0]);
		
		long ans = Long.MAX_VALUE;
		for(int i=0;i<n-k+1;i++) {
			for(int j=i+k-1;j<n;j++) {
				int[] y = new int[j-i+1];
				for(int l=0;l<j-i+1;l++) {
					y[l] = list.get(l+i)[1];
				}
				Arrays.sort(y);
				int yWidthMin = y[k-1] - y[0];
				for(int l=1;l+k-1<j-i+1;l++) {
					yWidthMin = Math.min(yWidthMin, y[l+k-1]-y[l]);
				}
				int xWidth = list.get(j)[0] - list.get(i)[0];
				ans = Math.min(ans, (long)xWidth*yWidthMin);
			}
		}
		
		System.out.println(ans);
		in.close();
		

	}

}