Ivan has number b. He is sorting through the numbers a from 1 to 10^{18}, and for every a writes ([a,    b])/(a) on blackboard. Here [a,    b] stands for least common multiple of a and b. Ivan is very lazy, that's why this task bored him soon. But he is interested in how many different numbers he would write on the board if he would finish the task. Help him to find the quantity of different numbers he would write on the board.

Input

The only line contains one integer — b (1 ≤ b ≤ 10^{10}).

Output

Print one number — answer for the problem.

Examples

Input

1


Output

1

Input

2


Output

2

Note

In the first example [a,    1] = a, therefore ([a,    b])/(a) is always equal to 1.

In the second example [a,    2] can be equal to a or 2 ⋅ a depending on parity of a. ([a,    b])/(a) can be equal to 1 and 2.
import java.io.*;
import java.util.StringTokenizer;

/**
 * Created by Anton Berezin on 2/18/2017.
 */
public class Nok {

	public static void main(String[] args) throws Exception {
		try (FastScanner fastScanner = new FastScanner();
		     PrintWriter writer = new PrintWriter(new BufferedOutputStream(System.out))) {
			long b = fastScanner.nextLong();
			writer.println(factors(b));

		}
	}

	private static long factors(long b) {
		if (b == 1) {
			return 1;
		}
		long result = 0;
		for (long i = 2; i * i <= b; i++) {
			if (b % i == 0) {
				result += i * i == b ? 1 : 2;
			}
		}
		return result + 2;
	}

	public static class FastScanner implements AutoCloseable {
		BufferedReader bufferedReader;
		StringTokenizer tokenizer;

		public FastScanner() {
			bufferedReader = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (tokenizer == null || !tokenizer.hasMoreElements()) {
				try {
					tokenizer = new StringTokenizer(bufferedReader.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return tokenizer.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = bufferedReader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		@Override
		public void close() throws IOException {
			bufferedReader.close();
		}
	}
}
