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:
/*
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/**
 *
 * @author ansam
 */
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        ArrayList<Integer> list = new ArrayList<>();
        int count = 0;
        for (int i = 1; count <= m; i++) {
            // the no. of zeros change every 5 numbers
            for (int j = i; j%5 == 0; j/=5) {
                count++;
                
            }
            if(count == m) list.add(i);
            
        }
        String x = "";
        System.out.println(list.size());
        for (Integer s : list) {
            System.out.print(x+s);
            x= " ";
        }
        System.out.println("");
    }
}