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: Runtime Error)

def gcd(a, b):
    while b != 0:
        a, b = b, a % b
    return a

def coprime(a, b):
    return gcd(a, b) == 1

def make_divisors(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)

    # divisors.sort()
    return divisors

A,B=map(int, input().split())
if (A==1)or(B==1)or(coprime(A,B)):
  print(1)
else:
  A=make_divisors(A)
  B=make_divisors(B)
  C=list(set(A)&set(B))
  C.sort(reverse=True)
  count=0
  for i in range(len(C)):
    for j in range(i+1,len(C)):
      if (C[j]!=1)and(C[i]%C[j]==0):
        count=count+1
        break
print(len(C)-count)