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 P633B
{
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		boolean none = false;
		int target = scan.nextInt();
		//Binary Search
		int lower = 0;
		int higher = 500000;
		int middle = 0;
		while (true)
		{
			if (higher - lower == 1)
			{
				if (zeros(lower) == target)
					middle = target;
				else if (zeros(higher) == target)
					middle = higher;
				else
					none = true;
				break;
			}
			middle = (lower+higher)/2;
			int z = zeros(middle);
			if (z < target) //too low
				lower = middle;
			else if (z > target) //too high
				higher = middle;
			else //got it
				break;
		}
		if (none)
			System.out.println(0);
		else
		{
			System.out.println(5);
			int first = middle - middle%5;
			for (int i = 0; i < 5; i++)
			{
				if (i != 0)
					System.out.print(" ");
				System.out.print(first + i);
			}
		}
		
		scan.close();
	}

	private static int zeros(int n)
	{
		//zeros = n/5 + n/25 + n/125 + ... + n/5^i
		int z = 0;
		for (int i = 1; i <= 13; i++)
			z += n/(int)Math.pow(5, i);
		return z;
	}
}