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 (4.02.3) (judge verdict: Runtime Error)

let filter n =
  match n with
  | 1 ->
      true
  | 2 ->
      true
  | n ->
      if n < 2 || n mod 2 = 0 then false
      else
        let rec f i =
          if float_of_int i > sqrt (float_of_int n) then true
          else if n mod i = 0 then false
          else f (i + 2)
        in
        f 3

let common_div a b =
  let sml = min a b and lrg = max a b in
  let num = Array.init sml (fun i -> i + 1) and acc = ref [] in
  for i = 0 to sml - 1 do
    if sml mod num.(i) = 0 && lrg mod num.(i) = 0 then acc := num.(i) :: !acc
  done ;
  Array.of_list @@ List.rev !acc

let () =
  let a, b = Scanf.scanf "%d %d\n" (fun a b -> (a, b)) in
  let divs = common_div a b in
  let c = ref 0 in
  Array.iter (fun n -> if filter n = true then c := !c + 1) divs ;
  print_int !c