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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class ATrivialProblem 
{
	
	public static int countFive(int n)
	{
		int count = 0;
		for (int i=5; n/i>=1; i *= 5)
	          count += n/i;
		
		return count;
	}
	
	public static int countTwo(int n)
	{
		int count = 0;
		for (int i=2; n/i>=1; i *= 2)
	          count += n/i;
		return count;
	}
	
	public static void main(String[] args) throws NumberFormatException, IOException 
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int m = Integer.parseInt(br.readLine());
		int count = 0;
		String result = "";
		for (int i = 1; i < 1000000; i++) 
		{
			int f =countFive(i);
			int t =countTwo(i);
			//System.out.println(f);
			//System.out.println(t);
			int zeros = Math.min(f,t);
			
			if(zeros == m)
			{
				count++;
				result += i +" ";
			}
			
			//if(zeros > m)
				//break;
		}
		System.out.println(count);
		if(count != 0)
		System.out.println(result);
		
		
	}
	

}