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.math.BigDecimal;
import java.text.DecimalFormat;
import java.util.*;

public class B {
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		Scanner s = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int m = s.nextInt();
		ArrayList<Integer> res = new ArrayList<>();
		for(int i = 1; i <= 10_000_000; i++) {
			int nt = numTrail(i);
			if(nt == m) res.add(i);
		}
		out.println(res.size());
		for(Integer i : res) out.print(i + " ");
		out.close();
	}
	
	static int numTrail(int n) {
		int res = 0;
		for(int i = 5; n/i >= 1; i *= 5) {
			res += n/i;
		}
		return res;
	}
}