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 (7.3.0) (judge verdict: Time Limit Exceeded)

def gcd(a,b):
  while b:
    a,b=b,a%b
  return a
a,b=map(int,input().split())
g=gcd(a,b)
def prime_table(n):
  n+=1
  table=[i if i%2 else 2 for i in range(n)]
  for i in range(3,n,2):
    if table[i]==i:
      for j in range(i*i,n,2*i):
        if table[j]==j:
          table[j]=i
  return table


def prime_fact(n,table):
  p=set()
  while n!=1:
    pp=table[n]
    p.add(pp)
    n//=pp
  return p

print(len(prime_fact(g,prime_table(g)))+1)