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.*;
public class Main {
    static int MAX = (int) 1e6;
    public static void main(String[] args) {
        IoSCANNER input = new IoSCANNER();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        ArrayList<Integer> res =new ArrayList<>();
        int n = input.nextInt();
        for(int i = 1;i <= MAX;i++ ){
            if(Count_TrailingZeros(i) == n )
                res.add(i);
        }
        out.println(res.size());
        if(res.size() != 0){
            for(int x=0;x<res.size();x++) out.print(res.get(x)+" ");
        }
        out.close();
    }

    public static int Count_TrailingZeros(int n){
        int total =0;
        for(int i =5;i<=n;i*=5) total+=(n/i);

        return total;
    }

}


class IoSCANNER{
    BufferedReader Br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    public String next() {
        while (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(Br.readLine());
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt() {return Integer.parseInt(next());}
    public long nextLong(){return  Long.parseLong(next());}
    public boolean hasNext(){
        try {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(Br.readLine());
        }
        catch (Exception e) {
            return false;
        }
        return true;

    }

}