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;
import java.util.Arrays;

public class Main{

static int MAX=400100;
    static int x[]=new int[MAX];
    public static void main(String args[]) throws IOException{

        for(int i=5;i<MAX;i+=5){
            x[i]=cal(i);
            x[i+1]=x[i];
            x[i+2]=x[i];
            x[i+3]=x[i];
            x[i+4]=x[i];
        }
        BufferedReader bufferedReader=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(bufferedReader.readLine());
        int index=Arrays.binarySearch(x,t);

        if(index<0)
            System.out.println("0");

        else {
            while(x[index-1]==x[index])
                index--;
            StringBuilder stringBuilder = new StringBuilder();
            stringBuilder.append("5\n");
            for (int i = 0; i < 5; i++)
                stringBuilder.append((index + i) + " ");
            System.out.println(stringBuilder.toString());
        }
    }

    private static int cal(int i) {
        int count=0;

         for(int j=1;j<=Math.log(i)/Math.log(5);j++){
             if(i/Math.pow(5,j)==0)
                 break;
             count+=i/Math.pow(5,j);
        }
        return count;
    }
}