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:
//package contest633;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Map;
import java.util.StringTokenizer;

/**
 * Created by Brijesh on 26-02-2016.
 */
public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        int m=Integer.parseInt(br.readLine());
        int n=5*(m+1),count=0;
        ArrayList<Integer> al=new ArrayList<>();
        for(int i=n;i>0;i--){
            int z=(int)(Math.floor(i/5)+Math.floor(i/25)+Math.floor(i/125)+Math.floor(i/625)+
                    Math.floor(i/3125)+Math.floor(i/15625)+Math.floor(i/78125)+Math.floor(i/390625));
            if(z<m)
                break;
            if(z==m){
                al.add(i);
                count++;
            }
        }
        System.out.println(count);
        Collections.sort(al);
        for(int i:al)
            System.out.print(i+" ");
    }
}