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.Scanner;

/**
 * http://codeforces.com/contest/633/problem/B
 */
public class B {

	public static void main(String[] args) {
		new B().doJob();
	}

	private void doJob() {
		try (Scanner scanner = new Scanner(System.in)) {
			int m = scanner.nextInt();
			int count = 0;
			StringBuilder builder = new StringBuilder();
			for (int i = 1; i < Integer.MAX_VALUE; i++) {
				int trailingZeros = countTrailingZeros(i);
				if (trailingZeros > m) {
					break;
				}
				if (trailingZeros == m) {
					count++;
					builder.append(i).append(" ");
				}
			}
			System.out.println(count);
			if (count > 0) {
				System.out.println(builder.toString().trim());
			}
		}
	}

	private int countTrailingZeros(int number) {
		int result = 0;
		for (int i = 1; number / Math.pow(5, i) >= 1; i++) {
			result += number / Math.pow(5, i);
		}
		return result;
	}
}