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: Runtime Error)

#!/usr/bin/env python3
import sys
import math


def solve(A: int, B: int):
    # al = make_divisors(A)
    # bl = make_divisors(B)
    # l = sorted(al & bl)
    gcd = math.gcd(A, B)
    f = factorize(gcd)
    print(1 + len(f))


def factorize(n: int):
    res = []
    end = int(n ** 0.5//1 + 2)
    for i in range(2, end):
        if n % i:
            continue
        s = 0
        while n % i == 0:
            n /= i
            s += 1
        res.append([i, s])
    return res


# def make_divisors(n):
#     divisors = []
#     for i in range(1, int(n**0.5)+1):
#         if n % i == 0:
#             divisors.append(i)
#             if i != n // i:
#                 divisors.append(n//i)

#     divisors.sort()
#     return divisors


# Generated by 1.1.5 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    A = int(next(tokens))  # type: int
    B = int(next(tokens))  # type: int
    solve(A, B)

if __name__ == '__main__':
    main()