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: Accepted)

module Main where

import qualified Data.ByteString.Char8 as BS
import Data.List
import Data.Char
import qualified Data.Map.Strict as M
-- import Debug.Trace

main :: IO ()
main = do
  [a, b] <- unfoldr (BS.readInteger . BS.dropWhile isSpace) <$> BS.getLine
  print . succ . M.size . M.filter (/= 0) . M.fromAscList . primeFactor $ gcd a b


-- http://qiita.com/myuon_myon/items/ad006568bd187223f494#2-4
primes = 2:3:[x|i<-[1..], j<-[-1,1], let x = 6*i+j, isPrime x] where
  isPrime n = null [i|i<-takeWhile (\x -> x*x <= n) primes, rem n i == 0]

-- | returns [(PrimeFactor, Exponent)]
-- >>> primeFactor 9
-- [(2,0),(3,2)]
primeFactor x = go x primes
  where       
    ub = floor . sqrt $ fromIntegral x
    go x (p:ps) 
      | p <= ub = let (x', e) = gogo x p 0 
        in (p, e) : go x' ps
      | x == 1 = []
      | otherwise = [(x, 1)]
      where
        gogo x p e = case quotRem x p of
          (q, 0) -> gogo q p (e + 1)
          (q, _) -> (x, e)