n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1

Is called the factorial of n. For example, the factorial of 12

12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 479001600

And there are two consecutive 0s at the end.

Write a program that inputs the integer n and outputs the number of consecutive 0s at the end of n !. However, n is a positive integer less than or equal to 20000.



Input

Multiple data are given. Each piece of data is given n (n ≤ 20000) on one line. When n is 0, it is the last input.

The number of data does not exceed 20.

Output

For each data, output the number of 0s that are consecutively arranged at the end of n! On one line.

Example

Input

2
12
10000
0


Output

0
2
2499
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Main {
	private	static	BufferedReader	br = null;

	static {
		br = new BufferedReader(new InputStreamReader(System.in));
	}

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int	num = 0;
		while((num = parseNum())!= 0) {
			fact(num);
		}
	}

	private static void fact(int n) {
		int		t = 0;
		int[]	a = new int[2];

		for(int i = 2; i <= n; i++) {
			for(t = i; t % 2 == 0; t /= 2) {
				a[0]++;
			}

			for(t = i; t % 5 == 0; t /= 5) {
				a[1]++;
			}
		}

		System.out.println(Math.min(a[0], a[1]));
	}

	private static int parseNum() {
		int		num    = 0;
		String	strNum = parseStdin();

		if (strNum != null) {
			num = Integer.parseInt(strNum);
		}

		return num;
	}

	private static String parseStdin() {
		String	strNum = null;

		try {
			String line = br.readLine();
			if (line != null) {
				if (!line.isEmpty()) {
					strNum = line;
				}
			}
		}
		catch (IOException e) {}

		return strNum;
	}
}