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.util.*;
import java.io.*;

public class B {

	static final boolean stdin = true;
	static final String filename = "";
	static FastScanner br;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException {

		if (stdin) {
			br = new FastScanner();
			pw = new PrintWriter(new OutputStreamWriter(System.out));
		} else {
			br = new FastScanner(filename + ".in");
			pw = new PrintWriter(new FileWriter(filename + ".out"));
		}

		Solver solver = new Solver();
		solver.solve(br, pw);
		pw.close();
	}

	static class Solver {
		static long mod = (long) (1e9);

		public void solve(FastScanner br, PrintWriter pw) throws IOException {
			long b = br.nl();
			long cnt = 0;
			for (long i = 1; i <= Math.sqrt(b); i++) {
				if (b % i == 0) {
					if (b / i == i) {
						cnt++;
					} else {
						cnt = cnt + 2;
					}
				}
			}
			System.out.println(cnt);
		}

		static long gcd(long a, long b) {
			if (a > b)
				return gcd(b, a);
			if (a == 0)
				return b;
			return gcd(b % a, a);
		}

		static long lcm(long a, long b) {
			return a * (b / gcd(a, b));
		}

		static long pow(long a, long b) {
			if (b == 0)
				return 1L;
			long val = pow(a, b / 2);
			if (b % 2 == 0)
				return val * val % mod;
			else
				return val * val % mod * a % mod;
		}

	}

	static class Point implements Comparable<Point> {
		int a;
		int b;

		Point(int a, int b) {
			this.a = a;
			this.b = b;
		}

		@Override
		public int compareTo(Point o) {
			return this.a - o.a;
		}

		public boolean equals(Object obj) {
			if (obj instanceof Point) {
				Point other = (Point) obj;
				return a == other.a && b == other.b;
			}
			return false;
		}

		public int hashCode() {
			return 65536 * a + b + 4733 * 0;
		}

	}

	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

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

		ArrayList<Integer>[] ng(int n, int e) {
			ArrayList<Integer>[] adj = new ArrayList[n];
			for (int i = 0; i < n; i++) {
				adj[i] = new ArrayList<Integer>();
			}
			for (int i = 0; i < e; i++) {
				int a = ni() - 1;
				int b = ni() - 1;
				adj[a].add(b);
				adj[b].add(a);
			}
			return adj;
		}

		Integer[] nIa(int n) {
			Integer[] arr = new Integer[n];
			for (int i = 0; i < n; i++) {
				arr[i] = ni();
			}
			return arr;
		}

		int[] nia(int n) {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = ni();
			}
			return arr;
		}

		Long[] nLa(int n) {
			Long[] arr = new Long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nl();
			}
			return arr;
		}

		long[] nla(int n) {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nl();
			}
			return arr;
		}

		String[] nsa(int n) {
			String[] arr = new String[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nt();
			}
			return arr;
		}

		String nt() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int ni() {
			return Integer.parseInt(nt());
		}

		long nl() {
			return Long.parseLong(nt());
		}

		double nd() {
			return Double.parseDouble(nt());
		}

	}
}