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.math.BigInteger;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int m = in.nextInt();
		int numof2 = 0;
		int numof5 = 0;
		int first = Integer.MAX_VALUE;
		int size = 0;
		for (int i = 2;; i++) {
			int temp = i;
			//System.out.println(temp);
			while (temp % 2 == 0 && temp > 0) {
				numof2++; temp /= 2;
			}
			while (temp % 5 == 0 && temp > 0) {
				numof5++; temp /= 5;
			}
			if (Math.min(numof2, numof5) == m) {
				first = Math.min(first, i);
				size++;
			}
			if (Math.min(numof2, numof5) > m)
				break;
			//System.out.println(i + " " + numof2 + " " + numof5);
		}
		System.out.println(size);
		for (int i = first; i < first + size; i++)
			System.out.print(i + " ");
	}
}