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

a_prime = a.prime_division.map{|x,y| x}
b_prime = b.prime_division.map{|x,y| x}
nums = a_prime & b_prime

ans = 0
d = nums.size
(2**d).times do |i|
  bits = i.to_s(2).split(//)
  (d - bits.length).times{bits.unshift('0')}

  selected = []
  bits.each_with_index do |bit, i|
    if bit == '1'
      selected.push nums[i]
    end
  end

  if selected.combination(2).all?{|x,y| x.gcd(y) == 1}
    # p ['selected', selected]
    ans = [ans, selected.count].max
  end

end

puts ans + 1