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

import math
def Sieve_of_Eratosthenes(N):
    if N<2: return set()
    prime=set(i for i in range(2,N+1))
    for i in range(2,math.ceil(N**0.5)+1):
        tmp=set()
        for j in prime:
            if j>i and j%i==0: tmp.add(j)
        for j in tmp:
            prime.remove(j)

    return prime

def gcd(a,b):
    if b>a:
        a,b=b,a
    while b:
        a,b=b, a%b
    return a

a,b=map(int,input().split())
n=gcd(a,b)
prime=list(Sieve_of_Eratosthenes(int(math.sqrt(n))+2))

ans=1
x=n
for p in prime:
    if x==1:
        break
    if x%p==0:
        ans+=1
        while x%p==0:
            x=x//p

if x!=1: ans+=1
print(ans)