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;

/**
 * Created by Артур on 27.02.2016.
 */
public class MainClass {
    public static void main(String[] args){
        int m=0;
        int x=0;
        int zeros=0;
        Scanner scanner = new Scanner(System.in);

        //System.out.print(logOf5(125));

        m=scanner.nextInt();

       // System.out.print(Math.log(25.0)/Math.log(5.0));

        while (zeros<m){
            x=x+5;
            zeros= (int) (zeros+valueOfZeros(x));
            //System.out.println(x+" "+zeros);
        }
        if (zeros==m){
            System.out.println(5);
            System.out.print(x+" ");
            for(int i=0; i<4; i++){
                x=x+1;
                System.out.print(x+" ");
            }
        } else {
            System.out.println(0);
        }
    }

    public static int valueOfZeros(int x){
        int otv=0;
        int pokazatel=0;
        int ost=0;
        while (x%5==0){
            x=x/5;
            otv++;
        }
        if (otv==0){
            otv=1;
        }
        return otv;
    }
}