Problem:
Score : 400 points

Problem Statement

Given are positive integers A and B.

Let us choose some number of positive common divisors of A and B.

Here, any two of the chosen divisors must be coprime.

At most, how many divisors can we choose?

Definition of common divisor

An integer d is said to be a common divisor of integers x and y when d divides both x and y.

Definition of being coprime

Integers x and y are said to be coprime when x and y have no positive common divisors other than 1.

Definition of dividing

An integer x is said to divide another integer y when there exists an integer \alpha such that y = \alpha x.

Constraints

All values in input are integers.

1 \leq A, B \leq 10^{12}

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of divisors that can be chosen to satisfy the condition.

Sample Input 1

12 18

Sample Output 1

3

12 and 18 have the following positive common divisors: 1, 2, 3, and 6.

1 and 2 are coprime, 2 and 3 are coprime, and 3 and 1 are coprime, so we can choose 1, 2, and 3, which achieve the maximum result.

Sample Input 2

420 660

Sample Output 2

4

Sample Input 3

1 2019

Sample Output 3

1

1 and 2019 have no positive common divisors other than 1.

Solution in OCaml (4.10.0) (judge verdict: Accepted)

Scanf.scanf "%d %d" (fun a b ->
    let rec gcd a b = if b = 0 then a else gcd b (a mod b) in

    let rec loop rest i acc =
        if rest = 1 then acc else
            if i * i > rest then acc + 1 else
                if rest mod i > 0 then loop rest (i + 1) acc else
                    let rec loop2 rest =
                        if rest mod i = 0 then loop2 (rest / i) else rest
                    in
                    loop (loop2 rest) (i + 1) (acc + 1)
    in
    loop (gcd a b) 2 1 |> Printf.printf "%d\n"
)