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)

import Data.List

main = do
  li <- getLine
  let [a,b] = map read $ words li
  let ans = compute a b
  print ans

-- 共通する素因数の個数を求める
compute :: Int -> Int -> Int
compute a b = succ $ length $ is (sub a) (sub b)

sub n = map head $ group $ primeFactors n

-- ordered intersection
is xxs@(x:xs) yys@(y:ys) =
  case compare x y of
    LT -> is xs yys
    EQ -> x : is xs ys
    GT -> is xxs ys
is _ _ = []

-- 自作の簡易素数と素因数分解
primes = 2 : 3 : 5 : sieve [(9,3),(10,5)] [7,9..]

{- (np,p)のリストと、奇数の列 -}
sieve ((q,p):qps) xxs@(x:xs) =
  case compare q x of
    LT -> sieve qps1 xxs
    EQ -> sieve qps1 xs
    GT -> x : sieve (insert (x+x,x) qps) xs
  where
    qps1 = insert (q+p,p) qps

primeFactors n = loop n primes
  where
    loop 1 _ = []
    loop n pps@(p:ps)
      | r == 0 = p : loop q pps
      | q < p  = [n]
      | True   = loop n ps
      where (q,r) = divMod n p