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)

# coding: utf-8
# Your code here!
A,B=map(int,input().split())

def gcd (a,b):
    temp=0
    M=max(a,b)
    m=min(a,b)
    for i in range(-int(-M**0.5//1)):
        if M%(i+1)==0:
            if m%(i+1)==0:
                temp=max(temp,(i+1))
            if m%(M//(i+1))==0:
                temp=max(temp,M//(i+1))
    return temp

koyaku=(gcd(A,B))

count=0
for i in range(2,-int(-koyaku**0.5)+1):
    while koyaku%i==0:
        koyaku//=i
        count+=1
print(count+1)