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)

def odd():
    n = 1
    while True:
        n = n + 2
        yield n


def divisible(n):
    return lambda x: x % n > 0


def primes():
    yield 2
    it = odd()
    while True:
        n = next(it)
        yield n
        it = filter(divisible(n), it)


count = 1
a, b = input().split(" ", 1)
a, b = int(a), int(b)
for x in primes():
    flag1 = 0
    flag2 = 0
    if x > a or x > b:
        break
    while a % x == 0:
        a /= x
        flag1 = 1
    while b % x == 0:
        b /= x
        flag2 = 1
    if flag1 == 1 and flag2 == 1:
        count += 1
print(count)