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 Ruby (2.3.3) (judge verdict: Accepted)

require 'prime'

class Integer
  def my_divisor_list2
    return [1] if self == 1
    Prime.prime_division(self).map do |e|
      Array.new(e[1]+1).map.with_index do |element, i|
        e[0]**i
      end
    end.inject{|p,q| p.product(q)}.map do |a|
      [a].flatten.inject(&:*)
    end.sort
  end
end

a, b = gets.split(' ').map(&:to_i)

as = a.my_divisor_list2
bs = b.my_divisor_list2

ks = as & bs

p ks.reject { |k|
  k != 1 && ks.any? { |k2| k2 != 1 && k != k2 && k % k2 == 0 }
}.length