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;

public class Fatorial {

	public static int howManyZeros (int fat, int m){
		int result = 0; 
		int soma = 0;
		boolean out = false;
		
		for (int i = 1; out != true; i++){
			result = (int)(fat/Math.pow(5,i));
			if (result < 1){
				return soma;
			}
			soma += result;
		}
		return soma;
	}
	
	public static int find (int m){
		int zeros = 0;
		boolean stop = false;
		
		if (m > 100000){
			return 0;
		}
		
		for (int i = 1; stop != true; i++) {	
			
			zeros = howManyZeros(i, m);
			
			if (zeros == m){
				stop = true;
				return i;
			}else if (zeros > 100000){
				return 0;
			}
			//System.out.println(i + " -> " + zeros);
		}
		return 0;		
	}
	
	public static void main (String[] args) {
		Scanner entrada = new Scanner(System.in);
		int m = entrada.nextInt();
		entrada.close();
		int result = 0;
		
		result = find(m);
		
		if (result != 0){
			System.out.println ("5");
			System.out.println (result + " " + (result + 1) + " " + (result + 2) + " " + (result + 3) + " " + (result + 4));
		}else{
			System.out.println ("0");
		}	
	}

}