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: Time Limit Exceeded)

import math as m
import fractions

A, B = list(map(int, input().split()))
#A , B = 12, 18

def sieve_of_eratosthenes(n):
    candidate = [1 for i in range(n+1)]
    candidate[0] = 0
    candidate[1] = 0
    for i in range(len(candidate)):
        if (candidate[i] == 0):
            continue
        j = 2
        while j * i < len(candidate):
            candidate[i * j] = 0
            j = j + 1
    primes = [i for i in range(n) if candidate[i] == 1]
    return primes

def prime_factorize(A):
    p = sieve_of_eratosthenes(m.floor(A**0.5))
    a = [0 for i in p]
    for i in range(len(p)):
        if A % p[i] != 0:
            a[i] = 0
        else:
            temp = A
            while temp % p[i] == 0:
                temp = temp / p[i]
                a[i] = a[i] + 1
    return a

#pf = prime_factorize(fractions.gcd(A, B))
# num = 1

# for x in pf:
#     if x >= 1:
#         num = num + 1



# print(num)

g = fractions.gcd(A,B)
primes = []
num = 1

if True:
    i = 2
    while i <= g:
        if (g % i == 0):
            cnt = 0
            while g % i == 0:
                cnt = cnt + 1
                g = g // i
            primes.append([i, cnt])
        i = i + 1


#print(primes)
print(len(primes)+1)