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.Scanner;

public class B {
	public static void main(String[] args) {
		
		Scanner s = new Scanner(System.in);
		int z = s.nextInt();
		//if z is power of 5
		int n = 1, ind = 0;
		boolean printed = false;
		ArrayList<Integer> a = new ArrayList<Integer>();
		while(true) {
			if(findTrailingZeros(n)==z) {
				a.add(n);
				printed = true;
			}
			if(findTrailingZeros(n)>z) break;
			n++;
		}
		if(!printed) System.out.println(0);
		else {
			System.out.println(a.size());
			for(int i = 0; i < a.size(); i++) System.out.print(a.get(i) + " ");
		}
	}
	
	static int findTrailingZeros(int  n)
	{
	    // Initialize result
	    int count = 0;
	 
	    // Keep dividing n by powers of 5 and update count
	    for (int i=5; n/i>=1; i *= 5)
	          count += n/i;
	 
	    return count;
	}
}