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)

import fractions
a,b=map(int,input().split())
g=fractions.gcd(a,b)

def make_eratosthenes(n):
    A=[0]*n
    A[0],A[1]=-1,-1 #0と1は対象外
    for i in range(2,n):
        if A[i]==0:
            j=i
            while j<n:
                A[j]=i
                j+=i
    return A

E=make_eratosthenes(g+100)
l={}
now=g
f=[] #素因数格納配列
while now>1:
    f.append(E[now])
    now//=E[now]

f=set(f)
print(len(f)+1)