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.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Scanner;

public class Cf633B {
    public static void main(String[] args) throws Exception {
        Scanner cin = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter cout = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int m = cin.nextInt();
        int k = 0;
        int cur = 0;
        while(k<m){
            cur+=5;
            k+=calc(cur);
        }
        if(k==m){
            cout.println(5);
            cout.println(cur+" "+(cur+1)+" "+(cur+2)+" "+(cur+3)+" "+(cur+4));
        }else{
            cout.println(0);
        }
        cout.close();
    }

    private static int calc(int cur){
        int rez = 0;
        while(cur>0 && cur%5==0){
            rez++;
            cur/=5;
        }
        return rez;
    }
}