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.*;
import java.util.*;
public class b {
public static void main(String[] args) throws IOException {
	input.init(System.in);
	PrintWriter out = new PrintWriter(System.out);
	int n = input.nextInt();
	int res = 0;
	int count = 0;
	while(count < n)
	{
		res++;
		int x = res;
		while(x%5 == 0)
		{
			x /= 5;
			count++;
		}
	}
	if(count == n)
	{
		out.println(5);
		for(int i = 0; i<5; i++) out.print(res+i+" ");
	}
	else out.println(0);
	out.close();
}

public static class input {
	static BufferedReader reader;
	static StringTokenizer tokenizer;

	static void init(InputStream input) {
		reader = new BufferedReader(new InputStreamReader(input));
		tokenizer = new StringTokenizer("");
	}

	static String next() throws IOException {
		while (!tokenizer.hasMoreTokens())
			tokenizer = new StringTokenizer(reader.readLine());
		return tokenizer.nextToken();
	}

	static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
}
}