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 PyPy3 (2.4.0) (judge verdict: Accepted)

# 1. 最大公約数を求める(ユークリッドの互除法)
# 2. 最大公約数を2と3以上の奇数で順次割っていくする（下から計算したら、素数判定は不要）
# 3. 2.の結果に含まれる数の種類の数＋１が答え

def main():
  a, b = sorted(list(map(int, input().split())), reverse = True) #aのほうを大きくする
  while b != 0: a, b = b, a % b #最大公約数を求める
  print(num_facto(a) + 1)  
  
def num_facto(n):
  ans = 0
  max_n = int(n ** 0.5) + 1
  
  if n == 1: return ans  
  for cursor in two_and_odds(max_n):
    if n % cursor == 0: ans += 1
    while n % cursor == 0: n = n // cursor
      
  if n != 1: ans += 1
  return ans

def two_and_odds(max_n):
  if max_n >= 2: yield 2
  for i in range(3, max_n, 2): yield i
    
if __name__ == "__main__":
  main()