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.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

public class Main {

	static Scanner sc = new Scanner(System.in);

	static int m;
	static long two = 0, five = 0, zeros = 0;
	static ArrayList<Integer> list = new ArrayList<>();
	
	public static void main(String[] args) {
		m = sc.nextInt();
		
		for (int i = 1; zeros <= m; i++) {
			if (i % 2 == 0) {
				int _i = i;
				while (_i % 2 == 0) {
					two++;
					_i /= 2;
				}
			}
			if (i % 5 == 0) {
				int _i = i;
				while (_i % 5 == 0) {
					five++;
					_i /= 5;
				}
			}
			
			long min = Math.min(two, five);
			zeros += min;
			two -= min;
			five -= min;
			
			if (zeros == m) list.add(i);
		}
		
		System.out.println(list.size());
		for (int i : list) System.out.print(i + " ");
	}

}