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.ArrayList;
import java.util.Scanner;

/**
 * Created by Mr.Algorithm on 6/26/2016.
 */
public class A_Trivial_Problem {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int m;
        m = Integer.parseInt(input.next());
        int cnz =0;
        ArrayList<Integer>nums = new ArrayList<>();

        for (int i = 4; cnz <= m; i++) {
            int ans = 0;
            int count = 1;
            int d = 5;
            while (d <=i) {
                ans += (i / d);
                count++;
                d = (int)Math.pow(5, count);
            }
            if (ans == m)nums.add(i);
            cnz = ans;
        }
        if (nums.size() == 0) System.out.println(0);
        else {
            System.out.println(String.valueOf(nums.size()));
            for ( int i = 0; i < nums.size(); i++) {
                System.out.print( nums.get(i) + " ");
            }
            System.out.println();
        }
    }
}