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)

def gcd(a,b)
  if b == 0
    a
  else
    gcd(b,a%b)
  end
end

def divided(x)
  ans = [1]
  range = (x ** 0.5).floor
  for i in 2..range
    once = false
    while x % i == 0
      x = x / i
      ans.push(i) if !once
      once = true
    end
  end
  ans
end

a, b = gets.chomp.split(" ").map(&:to_i)
gcd = gcd(a,b)
divisors = divided(gcd)
ans = 0
ans += divisors.length
divisors.each do |x|
  if divided(gcd/x).length == 1 && x!=1
    ans += 1
  end
end
puts ans