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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

/**
 * Created by brajesh.k on 26/02/16.
 */
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int m = Integer.parseInt(br.readLine());

        List<Long> ret = new ArrayList<>();

        long lo = 1;
        while (true) {
            int zero = getTrailingFacZero(lo);
            if (zero == m) {
                ret.add(lo);
            } else if (zero > m) {
                break;
            }
            lo++;
        }

        System.out.println(ret.size());
        if (ret.size() > 0) {
            for (int i = 0; i < ret.size(); i++) {
                if (i == 0) {
                    System.out.print(ret.get(i));
                } else {
                    System.out.print(" " + ret.get(i));
                }
            }
            System.out.println();
        }

    }

    private static int getTrailingFacZero(long n) {
        int ret = 0;
        long den = 5;
        while (n / den > 0) {
            ret += (n / den);
            den *= 5;
        }
        return ret;
    }
}