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;

public class JibnehArabieh {

	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt(), c = 1;
		ArrayList<Integer> a = new ArrayList<>();
		while(true){
			if(findtrailingzeros(c) == n) a.add(c);
			if(findtrailingzeros(c) > n) break;
			c++;
		}
		System.out.println(a.size());
		for(int i = 0 ; i < a.size() ; i++) System.out.print(a.get(i) + " ");
	}
	static int findtrailingzeros(int n){
		int count = 0;
		for(int i=5; n/i>=1; i *= 5) count += n/i;
		return count;
	}
}