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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class C633_B {

	public static void main(String[] args) throws NumberFormatException, IOException {
		InputStreamReader isr = new InputStreamReader(System.in);
		BufferedReader br = new BufferedReader(isr);

		int n = Integer.valueOf(br.readLine().trim());
		int cnt = 0;
		long i=0;
		boolean found = false;
		while(true){
			long t = i;
			while(t!= 0 && t%5==0){
				t /= 5;
				cnt++;
			}
			if(n == cnt){
				found = true;
				break;
			}
			if(cnt > n)
				break;
			i=i+5;
		}
		if(found){
			int c=1;
			StringBuffer sb = new StringBuffer();
			sb.append(i + " ");
			long j=i+1;
			while(j%5 != 0){
				sb.append(j + " ");
				j++;
				c++;
			}
			System.out.println(c);
			System.out.println(sb.toString().trim());
		} else {
			System.out.println("0");
		}

	}

}