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: Accepted)

def gcd(x, y)
  while true
    r = x % y
    break if r == 0
    x = y
    y = r
  end
  return y
end

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

count = 1
flag = false
while g % 2 == 0 do
  g /= 2
  flag = true
end
count += 1 if flag

sqr = Math.sqrt(g).to_i
3.step(sqr, 2) do |d|
  flag = false
  while g % d == 0 do
    g /= d
    flag = true
  end
  count += 1 if flag
end
count += 1 if g > 1
puts count