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: Wrong Answer)

from collections import Counter
from fractions import gcd
a, b = map(int, input().split())
n = int(max(a, b)**(1/2)) + 1
prime_num_cnt = [0]*(n + 1)
prime_num = []
for i in range(2, n+1):
    if prime_num_cnt[i] == 0:
        prime_num.append(i)
        tmp = i
        for j in range(1, n):
            if tmp*j > n:
                break
            prime_num_cnt[tmp*j] = 1


g = gcd(max(a, b), min(a, b))
cnt = Counter()
m = 2
while m*m < g:
    while g % m == 0:
        g //= m
        cnt[m] += 1
    m += 1

print(len(cnt) + 1)