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 CF
{
	public static void main(String[] args)
	{
		int[] powers = { 5, 25, 125, 625, 3125, 15625, 78125, 390625 };
		Scanner scan = new Scanner(System.in);
		int m = scan.nextInt();
		int zeros = 0, start = 0;
		while (zeros < m)
		{
			start += 5;
			zeros++;
			for (int i = 1; i < 8; i++)
			{
				if (start % powers[i] == 0)
					zeros++;
				else
					break;
			}
		}
		if (zeros == m)
		{
			System.out.println(5);
			System.out.print(start);
			for (int i = 1; i < 5; i++)
				System.out.print(" " + (start+i));
		}
		else
			System.out.println(0);
		
		scan.close();
	}
}