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
import random

def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)

    # divisors.sort()
    return divisors


def is_prime3(q, k=50):
    q = abs(q)
    # 計算するまでもなく判定できるものははじく
    if q == 2: return True
    if q < 2 or q & 1 == 0: return False

    # n-1=2^s*dとし（但しaは整数、dは奇数)、dを求める
    d = (q - 1) >> 1
    while d & 1 == 0:
        d >>= 1

    # 判定をk回繰り返す
    for i in range(k):
        a = random.randint(1, q - 1)
        t = d
        y = pow(a, t, q)
        # [0,s-1]の範囲すべてをチェック
        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())
gcd = math.gcd(a, b)
divisors = make_divisors(gcd)

ans = 0
for divisor in divisors:
    if is_prime3(divisor) is True:
        ans += 1

print(ans + 1)