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.Scanner;

public class code{
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);
        int n=in.nextInt(),count=0;
        int[] a = new int[1000];
        for(int i=1;true;i++){
            if(n_zero(i)==n){
                a[count]=i;
                count++;
            }
            if(n_zero(i)>n)     break;
        }
        System.out.println(count);
        for(int i=0;i<count;i++)    System.out.print(a[i]+" ");
    }
    public static long n_zero(int n){
        long z=0,b=5;
        while(b<=n){
            z+=n/b;
            b*=5;
        }
        return z;
    }
}