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 Task02 {

static long cntz(long n){
	long rez=0, tek=n;
	while (tek/5>0&&tek%5==0){tek/=5; rez++;}
	//rez--;
	return rez;
}
public static void main(String[] args) {
	Scanner input = new Scanner(System.in);
	long m = input.nextLong();
	long k=0, cnt=0;
	long rez=0;
		
	for (long i=5; k<=m; i+=5){
		//if (i%5==0) k++;
		//System.out.println(k);
		//if (i%25==0) k++;
		//System.out.println(k);
		k=k+cntz(i);
		//System.out.println(k);
		if (k==m){
		    rez=5;
			System.out.println(rez);
			System.out.format("%d %d %d %d %d", i, i+1, i+2, i+3, i+4);
			break;
		}
	}		
	if (rez==0)System.out.print(rez);
}
}