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();
      
      int[][] edges = new int[n][2];
      Set<Integer> xs = new HashSet<>();
      Set<Integer> ys = new HashSet<>();
      for(int i = 0; i < n; i++) {
        edges[i][0] = sc.nextInt();
        edges[i][1] = sc.nextInt();
        xs.add(edges[i][0]);
        ys.add(edges[i][1]);
      }

      long area = Long.MAX_VALUE;
      int cnt = 0;

      for(int xP : xs) {
          for(int yP: ys) {
              for(int xP2 : xs) {
                for(int yP2: ys) {
              
                  cnt = 0;
                  long sX = Math.min(xP, xP2), sY = Math.min(yP, yP2);
                  long eX = Math.max(xP, xP2), eY = Math.max(yP, yP2);
                  for(int m = 0; m < edges.length; m++) {
                      long x = edges[m][0], y = edges[m][1];
                      if(sX <= x && x <= eX && sY <= y && y <= eY)
                        cnt++;
                  }
                  if(cnt >= k)
                      area = Math.min(area, (long)(eX - sX) * (long)(eY - sY));
                }
              }
          }

      }

      System.out.println(area);
  }
}