Problem: Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

Solution (in JAVA):

Solution:
/*
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 * and open the template in the editor.
 */
import java.util.Scanner;
/**
 *
 * @author Great Blue Whale
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner x = new Scanner (System.in);
        int m  =  x.nextInt(),  number_of_0=0,number_of_multy_5=5, p = 0  ;
        long r [] = new long [100000]; 
        for(int i = 5; ((int)(i/number_of_multy_5)) <= m; i++) 
            {
                 for(;number_of_multy_5 <= i;)
                    {
                        number_of_0+=((int)(i/number_of_multy_5));
                        number_of_multy_5*=5;
                    }
                if(number_of_0 == m )r[p++]=i;
                number_of_0 = 0;
                number_of_multy_5 = 5;
            }
        if(r[0]==0)System.out.print(0);
        
        else 
            {
                System.out.print(p+"\n");
                for(int i = 0 ; i < p ; i++)
                    System.out.print(r[i]+ " ");
            }
    
    }
    
}