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: Runtime Error)

a, b = map(int, input().split())

# get GCD(Gratest Common Divisor: 最大公約数)
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

# get divisors(公約数)
def divisors(n):
    nums = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            nums.append(i)
            if i != n // i:
                nums.append(n//i)
    return nums.sort()

# is prime(Prime Number: 素数)
def is_prime(n):
    if n == 1: return False
    for k in range(2, int(n**0.5) + 1):
        if n % k == 0: return False
    return True

x = divisors(gcd(a, b))

ans = 1
for _x in x:
    ans += is_prime(_x)

print(ans)