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: Runtime Error)

A, B = gets.chomp.split.map(&:to_i)

def gcd(a, b)
  a, b = b, a if a < b

  while b > 0
    a, b = b, a % b
  end
  a
end

def fetch_prime_list(max)
  table = Array.new(max + 1,true)

  (2..max).each do |num|
    if table[num] == true
      k = num * num
      while k <= max
        table[k] = false
        k += num
      end
    end
  end
  return table
end

g = gcd(A, B)
p = []

1.upto(Math.sqrt(g).to_i) do |i|
  if g % i == 0
    p << i
    p << g / i
  end
end

is_prime = fetch_prime_list(g)

ans = 0

p.uniq.each do |num|
  ans += 1 if is_prime[num]
end

p ans