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: Accepted)

import fractions
import copy

a,b=map(int,input().split())

def calc_divisore(n):
    divisors=[]
    for i in range(1,int(n**0.5)+1):
        if n%i==0:
            divisors.append(i)
            if i!=n//i:
                divisors.append(n//i)
    return divisors

la=calc_divisore(a)
lb=calc_divisore(b)
ll=sorted(list(set(la)&set(lb)))
#print(la,lb,ll)
#print(ll)

ansl=[1]

ll_c=copy.copy(ll)

for i in ll_c:
#    ansl.append(i)
    len_s=len(ll)
    ll=list(filter(lambda x: fractions.gcd(i,x)==1, ll))
    if len(ll)!=len_s:
        ansl.append(i)

#   print(i,lll)
#    print(ansl)
#    print('\n')
print(len(ansl))