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 Trivial {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
			
		int zeros = 0;
		int i = 1;
		int c = 0;
		StringBuilder sb = new StringBuilder();
		while(zeros <= n) {
			if(i%5 == 0) {
				int m = i;
				while(m > 1 && m % 5 == 0) {
					zeros++;
					m = m /5;
				}
			}
			if(zeros == n) {
				c++;
				sb.append(i + " ");
			}
			i++;
		}
		if(c > 0) {
			System.out.println(c);
			System.out.println(sb.substring(0, sb.length()-1));
		} else {
			System.out.println(0);
		}
	}
	
}