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.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class CFB {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int m = in.nextInt();
		int fives = 0;
		int twos = 0;
		int tens = 0;
		int lam = 0;
		int laf = 0;
		int lat = 0;
		int i = 2;
		ArrayList<Integer> a = new ArrayList<Integer>();
		while (Math.min(fives, twos)<=m){
			int[] f = twoFiveFactor(i);
			twos+=f[0];
			fives+=f[1];
			if (Math.min(fives, twos)==m) a.add(i);
			i++;
		}
		System.out.println(a.size());
		for (Integer ii:a) System.out.print(ii+" ");
	}
	
	static int[] twoFiveFactor(int i){
		int[] f = new int[2];
		while (i%2==0){
			f[0]++;
			i/=2;
		}
		while (i%5==0){
			f[1]++;
			i/=5;
		}
		return f;
		
	}

}