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;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int K = scanner.nextInt();

    int[][] pts = new int[N][2];
    int[] x = new int[N];
    int[] y = new int[N];

    for (int i = 0; i < N; i++) {
      x[i] = scanner.nextInt();
      y[i] = scanner.nextInt();
      pts[i][0] = x[i];
      pts[i][1] = y[i];
    }

    Arrays.sort(x);
    Arrays.sort(y);

    long min = Long.MAX_VALUE;
    for (int i = 0; i < N; i++) {
      for (int j = i + 1; j < N; j++) {
        for (int k = 0; k < N; k++) {
          for (int l = k + 1; l < N; l++) {
            long left = x[i];
            long right = x[j];
            long bottom = y[k];
            long top = y[l];

            int count = 0;
            for (int m = 0; m < N; m++)
              if (pts[m][0] >= left
                  && pts[m][0] <= right
                  && pts[m][1] >= bottom
                  && pts[m][1] <= top) {
                count++;
              }
            if (count >= K) min = Math.min(min, Math.abs(right - left) * Math.abs(top - bottom));
          }
        }
      }
    }
    System.out.println(min);
  }
}