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)

import math

def factorization(n):
    arr = []
    tmp = n
    for div in range(2, int(-(-n**0.5//1))+1):
        if tmp % div == 0:
            cnt = 0
            while tmp % div == 0:
                cnt += 1
                tmp //= div
            arr.append([div, cnt])

    if tmp != 1:
        arr.append([tmp, 1])

    if arr == []:
        arr.append([n, 1])

    return arr

if __name__ == '__main__':
    A, B = map(int, input().split())
    gcd = math.gcd(A, B)
    if gcd == 1:
        print(1)
        exit()
    facts = factorization(gcd)
    print(len(facts) + 1)