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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import math
import fractions
a,b = list(map(int, input().split()))
c = fractions.gcd(a,b)

pn=[2]
for L in range(3,int(math.sqrt(c))+2):
    chk=True
    for L2 in pn:
        if L%L2 == 0:chk=False
    if chk==True:pn.append(L)

ans = 1
anslist = [1]
        
chk = True
for L2 in pn:
    if c%L2 == 0 : chk = False
if chk == True:
    anslist.append(c)
    ans+=1
#print(pn)


for i in range(len(pn)-1):
    if(c%pn[i]==0):
        ans+=1
        anslist.append(pn[i])
        chk = True
        for j in range(len(pn)-1):
            if((c//pn[i])%pn[j] == 0):
                chk = False
        if(chk == True):
            ans+=1
            anslist.append(c//pn[i])
print(ans)
#print(anslist)