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 LongContest1 {

	public static void main(String[] args) {
		/*for(int i=1;i<=500;i++) {
			System.out.println("trailing zero of "+i+" is "+trailingZero(i)+ " test "+i/5);
		}*/
		long n;
		Scanner sc = new Scanner(System.in);
		n = sc.nextLong();
		long lastindex = (n+1)*5;
		//System.out.println(lastindex);
		while(trailingZero(lastindex)>n) {
			//whatis("lastindex",lastindex);
			//whatis("trailingZero(lastindex)",trailingZero(lastindex));
			lastindex-=5;
		}
		long count = 0;
		lastindex+=5;
		while(trailingZero(lastindex) >= (n+1))
			lastindex--;
		if(trailingZero(lastindex)<n)
			System.out.println(0);
		else {
			System.out.println(5);
			long t=4;
			lastindex-=4;
			while(t-->=0) {
				System.out.print(lastindex+" ");
				lastindex++;
			}
			System.out.println();
		}
			
	}
	static void whatis(String varname, Object var) {
		System.out.println(varname+" is "+var.toString());
	}
	static long trailingZero(long n) {
		long count = 0;
		for(int i=5 ; n/i!=0;i*=5) {
			count+=n/i;
		}
		return count;
	}

}