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.List;
import java.util.Scanner;

/*
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/**
 *
 * @author Madi
 */
public class B633 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int k = 0;
        int p = 0;
        int q;
        boolean found = false;
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i <= 500000; i++) {
            p = 0;
            q = i;
            while (q > 0 && q % 5 == 0) {
                p++;
                q /= 5;
            }
            k += p;
            if (k == m) {
                list.add(i);
                found = true;
            } else if (k > m && !found) {
                System.out.println("0");
                return;
            }
        }
        
        if (found) {
            System.out.println(list.size());
            for (Integer i : list) {
                System.out.print(i + " ");
            }
            System.out.println();
        }
    }
}