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.*;
import java.util.*;
 
public class ClosingTweets {
	 
    public static void main(String[] args) throws IOException {
    	Scanner in=new Scanner(System.in);
    	//for(int i=1;i<1)
    	int[] a=new int[500000];
    	for(int i=1;i<500000;i++){
    		a[i]=tz(i);
    		//System.out.println(a[i]+" ");
    	}
    	
    	int m=in.nextInt();
    	int count=0;
    	int i=1;
    	while(a[i]!=m&&i<(500000-1)){
    		i++;
    	}
    	int x=i;
    	while(a[i]==m&&i<500000){
    		count++;
    		i++;
    	}
    	System.out.println(count);
    	//System.out.println();
    	for(int j=x;j<i;j++){
    		System.out.print(j+" ");
    	}
    	
    	//System.out.println(tz(10));
    	
    	
    	
    	
    	    	
    }
    
    public static int tz(int n) {
        long count = 0;
        for(long i= 5; n/i >= 1; i= i*5){
            count+= n/i;
        }
        return (int)count;
    }
    
}