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 b {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int m = scan.nextInt();
		int[] count = new int[100000];
		
		boolean start = false;
		int idx1 = -1, idx2 = -1;
		int t = 0;
		for(int i=1;t<=200000;i++){
			
			int k = 5*i;
			int c = 0;
			while(k%5==0){k/=5;c++;}
			t+= c;
			if(!start&&t==m){
				idx1=i*5;
				start=true;
			}
			if(start&&t!=m){
				idx2=i*5;
				break;
			}
			if(!start&&t>m)break;
		}
		
		System.out.println(idx2-idx1);
		if(idx2-idx1!=0){
			for(int i=idx1;i<idx2;i++)System.out.print(i+" ");
		}
	}
}