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.Scanner;


public class trivial {
	public static int numb(int a){
		int sum=0;
		int mul=5;
		while(true){
			if(a/mul == 0)
				return sum;
			sum += a/mul;
			mul *= 5;
				
		}
		
	}
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int eq;
		int n=sc.nextInt();
		eq=n;
		int m=0;
		int i;
		m=5*n;
		n= n*4;
		if(n%5==0){
			i=((m-n)/5)+1;
			for(int j=0;j<i;j++,n=n+5)
				if(numb(n)==eq){
					System.out.println(5);
					System.out.print(n+" "+(n+1)+" "+(n+2)+" "+(n+3)+" "+(n+4));
					System.exit(0);
			}
			System.out.println(0);}
		else{
			n= (n/5 +1)*5;
			i=((m-n)/5)+1;
			for(int j=0;j<i;j++,n=n+5)
				if(numb(n)==eq){
					System.out.println(5);
					System.out.print(n+" "+(n+1)+" "+(n+2)+" "+(n+3)+" "+(n+4));
					System.exit(0);
				}
			System.out.println(0);
			
			
			
		
		}
	}

}