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 (2.4.0) (judge verdict: Accepted)

#!/usr/bin/python3
# -*- coding: utf-8 -*-
from fractions import gcd


def main():
  A,B = map(int, input().split())
  AB = gcd(A,B)
  divs = set([])
  for i in range(1,int(AB**0.5)+1):
    if AB % i == 0:
      divs.add(i)
      divs.add(AB//i)
  divs = sorted(list(divs))

  rets = set([])
  for d in divs[1:]:
    chk = True
    for r in rets:
      if d%r == 0:
        chk = False
        continue
    if chk:
      rets.add(d)
  return len(rets)+1

if __name__ == "__main__":
  ret = main()
  print(ret)