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{
  static int n, k;
  static int[] x, y;
  static final long INF = Long.MAX_VALUE / 2;

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

    n = sc.nextInt();
    k = sc.nextInt();
    x = new int[n];
    y = new int[n];
    for(int i = 0; i < n; i++){
      x[i] = sc.nextInt();
      y[i] = sc.nextInt();
    }

    long min = INF;
    for(int i = 0; i < n; i++){
      for(int j = 0; j < n; j++){
        for(int kk = 0; kk < n; kk++){
          for(int l = 0; l < n; l++){
            int x1 = Math.min(x[i], x[j]);
            int x2 = Math.max(x[i], x[j]);
            int y1 = Math.min(y[kk], y[l]);
            int y2 = Math.max(y[kk], y[l]);
            int count = 0;
            for(int m = 0; m < n; m++){
              if(x1 <= x[m] && x[m] <= x2 &&
                y1 <= y[m] && y[m] <= y2){
                ++count;
              }
            }
            if(count >= k){
              min = Math.min(min, (long)(x2 - x1) * (long)(y2 - y1));
            }
          }
        }
      }
    }

    System.out.println(min);
  }
}