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)

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

if a > b:
    a, b  = b, a
if a == 1 or b == 1:
  print(1)
  exit()

ls = []
for i in range(1, int(a**(1/2)) + 1):
    if a % i == 0:
        ls.append(i)
        ls.append(int(a/i))

ls1 = []
for j in ls:
    if b % j == 0:
        ls1.append(j)

ans = []
for k in ls1:
    if k == 1 or k == 2 or k == 3:
      ans.append(k)
      continue
    for x in range(2, int(k**(1/2)) + 1):
        if k % x == 0:
            break
        if x == int(k**(1/2)):
            ans.append(k)

print(len(ans))