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: Wrong Answer)

from sys import stdin, setrecursionlimit


def gcd(x, y):
    if x % y == 0:
        return y
    return gcd(y, x % y)


def is_prime(n):
    if n == 1:
        return False
    for i in range(2, int(n ** 0.5) + 1):
        if n % i == 0:
            return False
    return True


def main():
    input = stdin.buffer.readline
    a, b = map(int, input().split())

    g = gcd(a, b)

    ans = 1

    for i in range(2, int(g ** 0.5) + 1):
        if g % i == 0:
            if is_prime(i):
                ans += 1
            if is_prime(g // i) and i != g // i:
                ans += 1

    print(ans)


if __name__ == "__main__":
    setrecursionlimit(10000)
    main()