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.io.*;
import java.util.*;
import java.text.*;
import java.math.*;

public class main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static PrintWriter pw = new PrintWriter(System.out);
    public static String line;
    public static StringTokenizer st;
    public static ArrayList<ArrayList<Integer>> adjList;

    public static void main(String[] args) throws Exception{
        int N = Integer.parseInt(br.readLine());
        int total = 0;

        for(int i = 5; i <= 500000; i+=5){
            int p = 5;
            while(i % p == 0){
                total++;
                p *= 5;
            }
            if(total == N){
                pw.print(5 + "\n");
                for(int j = i; j < i + 5; j++){
                    if(j != i) pw.print(" ");
                    pw.print(j);
                }
                break;
            } else if(total > N){
                pw.print(0);
                break;
            }
        }
        pw.print("\n");
        pw.close();
        br.close();
    }
}