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)
if a == b
    only = a
elsif a > b
    large = a
    small = b
else
    large = b
    small = a
end
while small > 0
    large, small = small, large%small
end
if large == 1
    puts 1
    exit
end

com_div =[]
1.upto(large) do |i|
    if small%i == 0 && large%i == 0
        com_div.push(i)
    end
end

1.upto(com_div.length-1) do |i|
    delete_list = []
    break if com_div[-1*i] == nil
    (i+1).upto(com_div.length-1) do |j|
        if com_div[j] % com_div[i] == 0
            delete_list.push(com_div[j])
        end
    end
    delete_list.each{|num| com_div.delete(num)}
end

p com_div.length