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.FileInputStream;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {
    static final double EPS         = 1e-9f;
    static final double TO_DEG      = 57.29577951;

    // UP, RIGHT, DOWN, LEFT, UPPER-RIGHT, LOWER-RIGHT, LOWER-LEFT, UPPER-LEFT
    static int[] dx = { -1, 0, 1, 0, -1, 1,  1, -1 };
    static int[] dy = {  0, 1, 0,-1,  1, 1, -1, -1 };

    // Represents all moves of a knight in a chessboard
    static int[] dxKnightMove = { -1, -2, -2, -1,  1,  2, 2, 1 };
    static int[] dyKnightMove = {  2,  1, -1, -2, -2, -1, 1, 2 };

    static final char WHITE = 0;
    static final char GRAY = 1;
    static final char BLACK = 2;

    static final int MAX = 100000;

    static long sqr(int x) { return x*x; }

    static int gcd(int a, int b) {
        while (b != 0) {
            b = b ^ a;
            a = a ^ b;
            b = b ^ a;
            a = a % b;
        }
        return a;
    }

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

    static final int MAX_PRIME = 1000000;

    static boolean[] flag = new boolean[MAX_PRIME + 7];
    static List<Integer> primes = new ArrayList<>();

    static class PowerPrime {
        long p;
        int a;  // p^a
        PowerPrime(long p, int a) { this.p = p; this.a = a; };
    }

    static void sieve() {
        int i = 2;
        while (i*i <= MAX_PRIME) {
            for (int j = i*i; j <= MAX_PRIME; j += i) flag[j] = true;
            while (flag[++i]);
        }
        primes.add(2);
        for (i = 3; i <= MAX_PRIME; i += 2) {
            if (flag[i] == false) primes.add(i);
        }
    }

    static List<PowerPrime> factorizeAsPower(long n) {
        if (n == 1) return Arrays.asList(new PowerPrime[] { new PowerPrime(1, 1) });

        List<PowerPrime> ret = new ArrayList<>();
        int root = (int)Math.sqrt(n);
        for (int i = 0; primes.get(i) <= root; i++) {
            int p = primes.get(i);
            int a = 0;
            while (n % p == 0) { n /= p; a++; }
            if (a > 0) ret.add(new PowerPrime(p, a));
        }

        if (n > 1) ret.add(new PowerPrime(n, 1));

        return ret;
    }

    static long solve(long n) {
        if (n == 1) return 1;
        List<PowerPrime> list = factorizeAsPower(n);
        long ret = 1;
        for (PowerPrime pp : list) {
            ret = ret * (pp.a + 1);
        }
        return ret;
    }

    public static void main(String[] args) {
        sieve();

        Scanner scanner = inputFromSystem();

        long b = scanner.nextLong();
        System.out.println(solve(b));
    }

    static Scanner inputFromFile() {
        try { return new Scanner(new FileInputStream("input.txt")); }
        catch (FileNotFoundException e) { e.printStackTrace(); }
        return null;
    }

    static Scanner inputFromSystem() {
        return new Scanner(System.in);
    }
}

