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).sort
if a==1
  p 1
  exit
end
s=[]
t=[]
f = true
(a ** 0.5).floor.times do |i|
  if a%(i+1)==0 && b%(i+1)==0
    t << a/(i+1)
    s.each do |ss|
#      p [i+1, ss, (i+1)%ss]
      if (i+1)%ss==0 && ss!=1
        f=false
        break
      end
    end
    if f
      s << (i+1)
    end
    f = true
  end
end
t.reverse.each do |tt|
  if b%tt==0
    s.each do |ss|
      if tt%ss==0 && ss!=1
        f=false
        break
      end
    end
    if f
      s << tt
    end
    f=true
  end
end
p s.size