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)

a, b = gets.split.map(&:to_i)

def div(x, y)
  if x % y == 0
    y
  else
    div(y, x % y)
  end
end

max_common_divisor = div([a, b].max, [a, b].min)

arr = []

(max_common_divisor/2).times do |i|
  arr.push(i*2 + 1) if ((a % (i*2 + 1) == 0) && (b % (i*2 + 1) == 0))
end

arr.push(2) if max_common_divisor >= 2

ans_arr = []
size = arr.size

arr.reverse_each.each_with_index do |v1, i|
  (size - 1 - i).times do |v2|
    ans_arr.push(v1) if div([v1, arr[v2]].max, [v1, arr[v2]].min) != 1
  end
end

puts arr.size - ans_arr.uniq.size