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.*;
import java.lang.*;
import java.io.*;
 
 
public class Main
{
	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
         final int MAX = 100000;
			int n = sc.nextInt();
			int[] arr = new int[MAX];
			int count=1;
			for(int i=1;i<MAX;i++) {
				if(i%5==0)count++;
				if(i%25==0)count++;
				if(i%125==0)count++;
				if(i%625==0)count++;
				if(i%3125==0)count++;
				if(i%15625==0)count++;
				if(i%78125==0)count++;
				arr[i] = count++;
				
			}
			int i=0;
			for(i=0;i<MAX;i++) {
				if(arr[i] == n)break;
			}
			if(i<MAX) {
				System.out.println("5");
				int start = i*5;
				for(int j=0;j<5;j++)System.out.print((start+j) + " ");
			}
			else System.out.println("0");

	}
}