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:
//package mashiro;
import java.io.*;
import java.util.Map;
import java.util.HashMap;
public class Main {
	final static int N = 500000+10;
	static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
	static int m;
	static int[] num = new int[N];
	public static int binary(int l,int r,int k){
		int res = 0;
		while(l <= r){
			int mid = (l + r)>>1;
			if(num[mid] >= k){
				res = mid;
				r = mid - 1;
			}
			else
				l = mid + 1;
		}
		return res;
	}
	public static void main(String[] args) throws IOException{
		num[0] = 0;
		for(int i = 1;i < N;i++){
			num[i] = num[i-1];
			int j = i;
			while(j % 5 == 0){
				num[i]++; j/=5;
			}
		}
		cin.nextToken(); m = (int)cin.nval;
		int pos = binary(0,N - 1,m);
		if(num[pos] != m){
			cout.println(0);
		}
		else{
			cout.println(5);
			for(int i = pos;i < pos + 5;i++)
				cout.print(i+" ");
			cout.println();
		}
		cout.flush();
	}
}