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: Time Limit Exceeded)

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

class Integer
  def my_divisor_list
    divisors = [1]
    primes = []
    Prime.prime_division(self).each do |prime|
      prime[1].times {primes << prime[0]}
    end

    1.upto(primes.size) do |i|
      primes.combination(i) do |prime|
        divisors << prime.inject{|a,b| a *= b}
      end
    end

    divisors.uniq!
    divisors.sort!
    return divisors
  rescue ZeroDivisionError
    return
  end
end

N = a.my_divisor_list
M = b.my_divisor_list

L = N & M

L2 = L.sort {|a, b| b <=> a }
require 'prime'

n = 1
L2.each do |h|
if Prime.prime?(h) == true
  n += 1
end
end

puts n