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 B
{
	public static void main(String[] args) throws Exception
	{
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int m = in.nextInt();
		
		long[] pow = new long[16];
		pow[0] = 1;
		for(int i = 1; i < pow.length; i++)
		{
			pow[i] = pow[i - 1] * 5;
		}
		
		int x;
		for(x = 5; m > 0; x += 5)
		{
			for(int y = pow.length - 1; y > 0; y--)
			{
				if(x % pow[y] == 0)
				{
					m -= y;
					break;
				}
			}
		}
		
		if(m == 0)
		{
			out.println(5);
			
			for(int z = x - 5; z < x; z++)
			{
				out.print(z);
				
				if(z < x - 1)
				{
					out.print(" ");
				}
			}
			
			out.println();
		}
		else
		{
			out.println(0);
		}
		
		out.close();
	}
}