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)

# 111
list = [int(item) for item in input().split()]

res_list = []
c=0

if list[0] >= list[1]:
    a = list[0]
    b = list[1]
else:
    a = list[1]
    b = list[0]
    
    
res_list.append(b)

if int(b/2) > 1:
    
    if b % 2 == 0:
        res_list.append(int(b/2))
    
    for i in range(1,int(b/2)):
        if b % i ==0:
            res_list.append(i)

res = []
for k in res_list:
    if a % k == 0:
        res.append(k)

new_res = []

for l in reversed(res):
    flag = 0
    cnt = 0
    for m in res:
        if l % m == 0:
            cnt += 1
    if cnt <=2:
        new_res.append(l)
            
print(len(new_res))