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.*;
import java.io.*;
import java.math.*;

public class Factorial {

	public static void solve(int m) {
		int cur = 0;
		int tot = 0;
		while (tot < m) {
			cur++;
			int c = cur;
			while (c % 10 == 0) {
				tot++;
				c = c / 10;
			}
			while (c % 5 == 0) {
				tot++;
				c = c / 5;
			}
		}
		if (tot > m) {
			System.out.println(0);
			return;
		}
		ArrayList<Integer> nums = new ArrayList<>();
		nums.add(cur);
		cur++;
		while (cur % 5 != 0 && cur % 10 != 0) {
			nums.add(cur);
			cur++;
		}
		System.out.println(nums.size());
		for (Integer i : nums) System.out.print(i + " ");
	}

	public static void main(String[] args) {
		try {
			InputStreamReader is = new InputStreamReader(System.in);
			BufferedReader br = new BufferedReader(is);
			solve(Integer.parseInt(br.readLine()) );
		} catch (Exception e) {e.printStackTrace();}
	}

}