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)

from collections import defaultdict
A, B = map(int, input().split())
ls = defaultdict(int)
ans = 0
for i in range(1, max(int(A**0.5)+1, int(B**0.5)+1)):
  if A%i==0 and B%i==0:
    if ls[i]==0:
      ls[i]=1
      for n in range(2,int(i**0.5)+1):
        if i%n==0:
          break
      else:
        ans += 1
  if A%i==0:
    j = A//i
    if B%j!=0:
      continue
  elif B%i==0:
    j = B//i
    if A%j!=0:
      continue
  else:
    continue
  if ls[j]==0:
    ls[j] = 1
  else:
    continue
  for n in range(2,int(j**0.5)+1):
    if j%n==0:
      break
  else:
    ans += 1
print(ans)