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;
import java.util.function.Function;

/**
 * Created by btk on 2017/10/13.
 */
public class Main {
   public static void main(String[] args){
       Scanner sc=new Scanner(System.in);
       int n=sc.nextInt();
       int k=sc.nextInt();
       long[] x = new long[n];
       long[] y = new long[n];
       for (int i = 0; i < n; i++) {
           x[i]=sc.nextLong();
           y[i]=sc.nextLong();
       }
       long ret=4123456789123456789l;
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               for (int l = 0; l < n; l++) {
                   long lb=y[l];
                   long ub=1123456789;
                   final long xl=x[i];
                   final long xu=x[j];
                   final long yl=y[l];
                   Function<Long,Integer> calc=(m)->{
                       int sum=0;
                       for (int o = 0; o < n; o++) {
                           if(xl<=x[o]&&x[o]<=xu&&yl<=y[o]&&y[o]<=m){
                               sum++;
                           }
                       }
                       return sum;
                   };
                   if(calc.apply(ub)<k)continue;
                   while(ub-lb>1){
                       long m=(lb+ub)/2;
                       if(calc.apply(m)>=k)ub=m;
                       else lb=m;
                   }
                   ret=Math.min(ret,(xu-xl)*(ub-yl));

               }
           }
       }
       System.out.println(ret);
       sc.close();
   }
}