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.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

/**
 * 解説どおりに実装したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int k = scanner.nextInt();
			int[] x = new int[n], y = new int[n];
			IntStream.range(0, n).forEach(i -> {
				x[i] = scanner.nextInt();
				y[i] = scanner.nextInt();
			});
			long answer = Long.MAX_VALUE;
			int[] sortedX = x.clone();
			int[] sortedY = y.clone();
			Arrays.sort(sortedX);
			Arrays.sort(sortedY);
			for (int xi = 0; xi < n; xi++) {
				for (int xj = xi + 1; xj < n; xj++) {
					for (int yi = 0; yi < n; yi++) {
						for (int yj = yi + 1; yj < n; yj++) {
							int count = 0;
							for (int i = 0; i < n; i++) {
								if ((sortedX[xi] <= x[i]) && (x[i] <= sortedX[xj]) && (sortedY[yi] <= y[i])
										&& (y[i] <= sortedY[yj])) {
									count++;
								}
							}
							if (count >= k) {
								answer = Math.min(answer,
										1L * (sortedX[xj] - sortedX[xi]) * (sortedY[yj] - sortedY[yi]));
							}
						}
					}
				}
			}
			System.out.println(answer);
		}
	}
}