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.*;
import java.util.*;
public class B {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String word;
		while((word = br.readLine()) != null) {
			int n = Integer.parseInt(word);
			ArrayList<Integer> ans = new ArrayList<Integer>();
			ans = getAns(n);
			System.out.println(ans.size());
			if(ans.size() != 0) {
				for(int i = 0; i < ans.size()-1; i++) 
					System.out.print(ans.get(i) + " ");
				System.out.println(ans.get(ans.size()-1));
			}
		}
		
	}
	
	public static ArrayList<Integer> getAns(int n)  {
		ArrayList<Integer> r = new ArrayList<Integer>();
		int i = 0;
		int zeros = 0;
		while(zeros <= n) {
			int aux = i;
			while(aux > 0 && aux % 5 == 0) {
				zeros++;
				aux /= 5;
			}
			if(zeros == n) r.add(i);
			i++;
		}
		return r;
	}
}