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 Python (3.4.3) (judge verdict: Wrong Answer)

def isPrime(n):
  if n < 2:
    # 2未満は素数でない
    return False
  if n == 2:
    # 2は素数
    return True
  for p in range(2, n):
      if n % p == 0:
        # nまでの数で割り切れたら素数ではない
        return 0
  # nまでの数で割り切れなかったら素数
  return 1


def gcd(a, b):
    while b:
        a, b = b, a % b
    return a


def divisor(n):
    i = 1
    table = []
    while i * i <= n:
        if n % i == 0:
            table.append(i)
            table.append(n//i)
        i += 1
    table = list(set(table))
    return table

def main():
    a, b = map(int, input().split())
    g = gcd(a, b)
    d = divisor(g)
    print(d)
    cnt = 0
    for num in d:
        cnt += isPrime(num)
    print(cnt + 1)

if __name__ == "__main__":
    main()