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: Accepted)

a, b = [int(i) for i in input().split()]
small = min(a, b)
big = max(a, b)

import sys

if small == 1:
    print(1)
    sys.exit(0)

import math
import collections

# FYI: https://ja.wikipedia.org/wiki/%E8%A9%A6%E3%81%97%E5%89%B2%E3%82%8A%E6%B3%95
def prime_factorization(num, divisors=None):
    if num <= 1:
        assert False
    factors = []

    # divisorsを使い回せるようにする。
    #   - 二度生成した場合にかかる時間が許容できない場合
    #   - エラトステネスの篩などで素数テーブルを既に持っている場合
    if not divisors:
        divisors = [2] + [i for i in range(3, int(math.sqrt(num)) + 1) if i % 2 != 0]

    for d in divisors:
        # divisorsを引数で与えた場合のみ起こりうる
        if num < d:
            break
        while num % d == 0:
            factors.append(d)
            num //= d

    if num != 1:
        factors.append(num)
    return factors

divisors = [2] + [i for i in range(3, int(math.sqrt(big)) + 1) if i % 2 != 0]
small_factors = set(prime_factorization(small, divisors))
big_factors = set(prime_factorization(big, divisors))

print(len(small_factors.intersection(big_factors)) + 1)