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.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt(), k = sc.nextInt();
		
		int x[] = new int[n];
		int y[] = new int[n];
		
		int cx[] = new int[n];
		int cy[] = new int[n];
		
		for(int i=0;i<n;i++){
			x[i] = sc.nextInt();
			y[i] = sc.nextInt();
			cx[i] = x[i];
			cy[i] = y[i];
			
		}
		
		Arrays.sort(x);
		Arrays.sort(y);
		
		long ans = Long.MAX_VALUE;

		for(int lx=0;lx<n;lx++){ for(int rx=lx+1;rx<n;rx++){
			for(int ly=0;ly<n;ly++){for(int ry=ly+1;ry<n;ry++){
				int x1 = x[lx], x2=x[rx];
				int y1 = y[ly], y2=y[ry];
				
				int cnt = 0;
				for(int i=0;i<n;i++){
					if(x1 <= cx[i] && cx[i] <= x2 && y1 <= cy[i] && cy[i] <= y2){
						cnt++;
					}
				}
				if(cnt >= k){
					ans = Math.min(ans, (long)(x2-x1) * (y2-y1));
				}
			}
			}
		}
		}
		System.out.println(ans);
	}

}