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: Wrong Answer)

input = gets.chomp.split(" ").map(&:to_i).sort
 
require 'prime'
 
if input[0].prime? || input[1].prime?
  if input[0]==input[1]
    puts 2
  else
    puts 1
  end
else
  i = 2
  result = []
  count = 1
 
  while (i <= input[0]) do
    if input[0].prime?
      result << input[0]
      break
    end
    if !i.prime?
      break
    end
    if (input[0]%i ==0)
      result << i
      input[0] = input[0]/i
    end
    i += 1
  end
 
  result.each do |e|
    if input[1]%e ==0
      count += 1
    end
  end
 
  puts count
 
end