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)

def dmod(e,n)
  ( 43 * e + 1 ) % n
end

def gcd(a,b)
  r = a % b
  while r != 0
    a = b
    b = r
    r = a % b
  end
  b
end

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

if a < b
  z = a
  a = b
  b = z
end

b = gcd(a,b)
#p b

n = b
i = 1
ans = 1

x = 2
y = 2
visited = [1]

1000.times do
  x = dmod(x,n)
  y = dmod(dmod(y,n),n)
  d = gcd((x-y).abs,n)
  i += 1
  if d == 1
    next
  elsif d < n
    if !visited.include?(d)
      visited << d
      ans += 1
      #p d
    end
  else
    break
  end
end

p ans