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 a = sc.nextInt();
		int[] x = new int[n];
		int[] y = new int[n];
		for(int i = 0 ; i < n ; i++) {
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
		}
		long ans = 1L<<62;
		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++) {
						long cnt = 0;
						for(int m = 0 ; m < n ; m++) {
							if(x[i] <= x[m] && x[m] <= x[j] && y[k] <= y[m] && y[m] <= y[l]) {
								cnt++;
							}
						}
						if(cnt >= a) {
							ans = Math.min(ans, 1L * (x[j] - x[i]) * (y[l] - y[k]));
						}
					}
				}
			}
		}
		System.out.println(ans);
	}
}