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.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;

/**
 * Created by timur on 28.03.15.
 */

public class TaskB {
    boolean eof;
    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;

    public static void main(String[] args) throws IOException {
        new TaskB().run();
    }

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return "-1";
            }
        }
        return st.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

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

    String nextLine() throws IOException {
        return br.readLine();
    }


    void run() throws IOException {
        InputStream input = System.in;
        PrintStream output = System.out;
        try {
            File f = new File("a.in");
            if (f.exists() && f.canRead()) {
                input = new FileInputStream(f);
                output = new PrintStream("a.out");
            }
        } catch (Throwable e) {
        }
        br = new BufferedReader(new InputStreamReader(input));
        out = new PrintWriter(output);
        solve();
        br.close();
        out.close();
    }

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


    void solve() {
        boolean[] np;
        long[] primes;
        int MAXP = 100005;

        np = new boolean[MAXP];
        for (int i = 2; i < MAXP; i++) {
            if (!np[i]) {
                for (int j = i * i; j < MAXP && j > 0; j += i) {
                    np[j] = true;
                }
            }
        }
        int count = 0;
        for (int i = 2; i < MAXP; i++) {
            if (!np[i]) {
                count++;
            }
        }
        count++;
        primes = new long[count];
        for (int i = 2, j = 0; i < MAXP; i++) {
            if (!np[i]) {
                primes[j++] = i;
            }
        }
        long ans = 1, cm;
        long b = nextLong();
        for (int i = 0; i < count - 1; i++) {
            cm = 1;
            while (b % primes[i] == 0) {
                cm++;
                b /= primes[i];
            }
            ans *= cm;
        }
        if (b > 1) {
            ans *= 2;
        }
        out.print(ans);
    }
}
