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.math.BigInteger;
import java.util.Scanner;

/*
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/**
 *
 * @author root
 */
public class MtrailingZerosInFactorial {
 
    public static void main(String args[]){
        
        Scanner in = new Scanner(System.in);
        int m = in.nextInt();
        int zeros[] = new int[(m>0)?m+1:2];
        zeros[0] = 0;
        zeros[1] = 1;
        for(int i = 2;i<m+1;i++){
            int k = i;
            int x = 1;
            while(k%5 == 0){
                k/=5;
                x++;
            }
            zeros[i] = zeros[i-1]+x;
        }
        int index = -1;
//        System.out.println("dp:");
        for(int i = 0;i<m+1;i++){
//            System.out.print(zeros[i]+" ");
            if(zeros[i] == m){
                index = i;
                break;
            }
        }
        if(index>=0){
            System.out.println("5");
            int startIndex = (index)*5;
            for(int i = startIndex;i<startIndex+5;i++){
                System.out.print(i+" ");
            }
        } else {
            System.out.println("0");
        }
    }
//    
    public static BigInteger factorial(int a){
        BigInteger [] f = new BigInteger[a+1];
        f[0] = new BigInteger("1");
        f[1] = new BigInteger("1");
        for(int i = 2;i<=a;i++){
            f[i] = new BigInteger(i+"").multiply(f[i-1]);
        }
        
        return f[a];
    }
}