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();
		int K = sc.nextInt();
		long x[] = new long[N];
		long y[] = new long[N];
		for(int i = 0; i < N; i++) {
			x[i] = sc.nextLong();
			y[i] = sc.nextLong();
		}
		
		long ans = Long.MAX_VALUE;
		for(int left = 0; left < N; left++) {
			for(int right = 0; right < N; right++) {
				for(int up = 0; up < N; up++) {
					for(int down = 0; down < N; down++) {
						int cnt = 0;
						for(int i = 0; i < N; i++) {
							if(x[left] <= x[i] && x[i] <= x[right] && y[i] <= y[up] && y[down] <= y[i]) {
								cnt++;
							}
						}
						if(cnt >= K) {
							ans = Math.min(ans, Math.abs(x[left] - x[right]) * Math.abs(y[up] - y[down]));
						}
					}
				}
			}
		}
		System.out.println(ans);
	}
}