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)
$gcd = A.gcd(B)

def divisor(num)
    result = []
    i = 1
    while i <= $gcd do
        remainder = num % i
        if remainder == 0
            result << i
        end
        i += 1
    end 
    return result
end

d1 = divisor(A < B ? A : B)
commons = []
d1.each do |d|
    if (A > B ? A : B) % d == 0
        commons << d
    end
end
commons.delete(1)
commons.size.times do |i|
    k = i + 1
    while k < commons.size do
        if commons[k] % commons[i] == 0
            commons.delete(commons[k])
        end
        k += 1
    end
end


puts commons.size + 1