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:
import java.util.* ;
import java.io.*;
import java.math.BigInteger ;











public class Main {

        
    public static void main(String[] args) throws Exception{
    
 // BufferedReader bf = new BufferedReader( new InputStreamReader(System.in)) ;
   Scanner s = new Scanner(System.in) ;
   //   StringTokenizer st = new StringTokenizer(bf.readLine());
   int n = s.nextInt() ; StringBuilder an = new StringBuilder() ;
   
   long count = 0  , co = 0;
   for(long i = 5 ; ; )
   {
   count += 5 ;
  
   
   n-- ;
   
   co = 25 ;
   while(count >= co)
   {
       if(count % co == 0)
           n-- ;
       co *= 5 ;
   
   }
   
   if(n < 0){
           System.out.println("0"); System.exit(0); }
   if(n == 0)
       break ;
   }
        System.out.println("5");
        for(long i = count ; i < count + 5 ; i++)
        System.out.print(i + " ");
    }
    }