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 (2.7.1) (judge verdict: Accepted)

require 'set'

class Hash
  def push(key, value)
    self[key] = [] if self[key] == nil
    self[key].push(value)
  end
end

class Array
  def lower_bound(value)
    left = -1;
    right = self.length;
    while left + 1 < right
      mid = left + (right - left) / 2;
      if self[mid] >= value
        right = mid
      else
        left = mid
      end
    end
    right
  end

  def unique
    res = [self.first]
    each_cons(2) do |a, b|
      if a != b
        res.push(b)
      end
    end
    res
  end
end

def get_ints
  gets.split.map(&:to_i)
end

def get_ints_minus_one
  get_ints.map { |x| x - 1 }
end

def get_int
  gets.chomp.to_i
end

require 'prime'

A, B = get_ints

def divisor(n)
  result = SortedSet.new
  result.add(1); result.add(n)
  i = 1
  num = Math.sqrt(n)
  while i <= num do
    remainder = n % i
    if remainder == 0
      result.add(i)
      result.add(n/i)
    end
    i += 1
  end
  result.to_a
end

a_divisors = divisor(A)
b_divisors = divisor(B)

ab_divisors = (a_divisors & b_divisors).sort[1..] # 1 は必ず入っているので削除

count = 1 # 1を入れるので1からカウント
set = Set.new
i = 0
while i < ab_divisors.length
  unless set.include?(ab_divisors[i])
    count += 1
    j = i + 1
    while j < ab_divisors.length
      set.add(ab_divisors[j]) if ab_divisors[j] % ab_divisors[i] == 0
      j += 1
    end
  end
  i += 1
end

puts count