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)

require 'prime'
a, b = gets.split.map(&:to_i)

divs = Array.new(0)
for i in 1..10**6
  break if a < i && b < i
  divs.push(i) if a % i == 0 && b % i == 0
  divs.push(a/i) if a % i == 0 && b % (a/i) == 0
  divs.push(b/i) if b % i == 0 && a % (b/i) == 0
end
divs.uniq!
#puts divs.join(" ")
cnt = 0
for i in 0..(divs.length-1)
  cnt += 1 if Prime.prime?(divs[i]) || divs[i] == 1
end
puts cnt