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)

def slove():
    import sys
    input = sys.stdin.readline
    a, b = list(map(int, input().rstrip('\n').split()))
    ls = [True] * (int(max(a, b) ** 0.5) + 1)
    ls[0] = False
    ls[1] = False
    cnt = 1
    for i in range(len(ls)):
        if ls[i]:
            ls[i] = False
            if a % i == 0 and b & i == 0:
                cnt += 1
            for j in range(i, len(ls), i):
                if ls[j]:
                    ls[j] = False
    print(cnt)


if __name__ == '__main__':
    slove()