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 Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

import Data.List

factors :: Int -> [Int]
factors n = [x | x <- [1..n], n `mod` x == 0]

factorization :: Int -> [Int]
factorization 1 = []
factorization n
  | factors == []  = [n]
  | otherwise = factors ++ factorization (n `div` (head factors))
  where factors = take 1 $ filter (\x -> (n `mod` x) == 0) [2 .. n-1]

main :: IO ()
main = do
  [a, b] <- map read . words <$> getLine
  print . (+1) . length . group . factorization $ gcd a b