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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
     
    public class B {
    	
    	public static void main(String[] args) throws IOException{
     
    		Scanner sc = new Scanner(System.in);
    		PrintWriter out = new PrintWriter(System.out);
    		int n = sc.nextInt();
    		int cur = 5, count = 0;
    		while(true)
    		{
    			int tmp = cur; while(tmp % 5 == 0){tmp /= 5; count++;}
    			if(count >= n) break;
    			cur += 5;
    		}
    		if(count > n) out.println(0);	
    		else
    		{
    			out.println(5);
    			for(int i = cur; i < cur + 5; i++)
    				out.print(i + " ");
    			out.println();
    		}
    		out.flush();
    	}
    	
    	static class Scanner 
    	{
    		StringTokenizer st;
    		BufferedReader br;
     
    		public Scanner(InputStream System){	br = new BufferedReader(new InputStreamReader(System));	}
    		public String next() throws IOException 
    		{
    			while (st == null || !st.hasMoreTokens()) 
    				st = new StringTokenizer(br.readLine());
    			return st.nextToken();
    		}
    		public String nextLine()throws IOException{return br.readLine();}
    		public int nextInt() throws IOException {return Integer.parseInt(next());}
    		public double nextDouble() throws IOException {return Double.parseDouble(next());}
    		public char nextChar()throws IOException{return next().charAt(0);}
    		public Long nextLong()throws IOException{return Long.parseLong(next());}
    		public boolean ready() throws IOException{return br.ready();}
    		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
    	}
    }