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();
    ArrayList<Long> xlist = new ArrayList<Long>();
    ArrayList<Long> ylist = new ArrayList<Long>();
    long[][] co = new long[n][2];
    for(int i = 0; i < n; i++) {
      long x = sc.nextLong();
      long y = sc.nextLong();
      xlist.add(x);
      ylist.add(y);
      co[i][0] = x;
      co[i][1] = y;
    }
    Collections.sort(xlist);
    Collections.sort(ylist);
    long ans = Long.MAX_VALUE;
    for(int i = 0; i < n; i++) {
      for(int j = i + 1; j < n; j++) {
        for(int s = 0; s < n; s++) {
          for(int l = s + 1; l < n; l++) {
            long xmin = xlist.get(i);
            long xmax = xlist.get(j);
            long ymin = ylist.get(s);
            long ymax = ylist.get(l);
            int count = 0;
            for(int m = 0; m < n; m++) {
              if((co[m][0] >= xmin) && (co[m][0] <= xmax) && (co[m][1] >= ymin) && (co[m][1] <= ymax)) count++;
            }
            if(count >= k) ans = Math.min(ans, (xmax - xmin) * (ymax - ymin));
          }
        }
      }
    }
    System.out.println(ans);
  }
}