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)

A, B = gets.split(' ').map{|n| n.to_i}
cds = [1]
sum = 1
(2..([A, B].min)).each do |i|
  next if A % i != 0 || B % i != 0
  
  if cds.length == 1
    cds << i
    sum += 1
    next
  end
  
  catch :done do
    cds.each do |cd|
      if cd == 1
        next
      elsif i % cd == 0
        throw :done
      else        
        j = i % cd
        while j > 0 do
          cd, j = j, cd % j
        end
        throw :done if cd != 1
      end
    end
    cds << i
    sum += 1
  end
end
p sum