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 B {

	static int fives(int n) {
		int answer = 0;
		while (n >= 5) {
			if ((n % 5) == 0) {
				answer++;
				n/=5;
			}
			else
				break;
		}
		return answer;
	}

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String s = br.readLine();
		int n = Integer.parseInt(s);
		boolean notfound = false;
		int i = -5;
		int fives = 0;
		for (int h = 0; h < 100000 && fives <= 100000; h++) {
//			System.out.println(i);
			i += 5;
			fives += fives(i);
			if (fives == n)
				break;
			if (fives > n)
				notfound = true;
		}
//		System.out.println(fives(9));
		if (notfound)
			System.out.println(0);
		else {
			System.out.println(5);
			for (int k = 0; k < 5; k++)
				System.out.print(i + k + " ");
		}
	}

}