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: Wrong Answer)

import math
from math import ceil
import random

def is_prime(q,k=10000):
    q = abs(q)
    if q == 2 or q == 1: return True
    if q < 2 or q&1 == 0: return False

    d = (q-1)>>1
    while d&1 == 0:
        d >>= 1
    for i in range(k):
        a = random.randint(1,q-1)
        t = d
        y = pow(a,t,q)
        while t != q-1 and y != 1 and y != q-1:
            y = pow(y,2,q)
            t <<= 1
        if y != q-1 and t&1 == 0:
            return False
    return True


a, b = map(int, input().split())
yakusu = []
for i in range(1, ceil(math.sqrt(max(a, b)))+1):
    if a % i == 0 and b % i == 0 and is_prime(i):
        yakusu.append(i)
print(len(yakusu))