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 C++14 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <iostream>
#include <vector>

bool isPrime(long long readVal){
    switch(readVal){
        case 0:
            return false;
        case 1:
            return false;
        case 2:
            return true;
    }

    if(0 == readVal % 2){
        return false;
    }

    for(int i = 3; i * i <= readVal; ++i){
        if(0 == readVal % i){
            return false;
        }
    }

    return true;
}

long long gcd(long long a, long long b){
    if(0 == a % b){
        return a;
    }else{
        return gcd(b, b % a);
    }
}

int main(){
    long long a, b;
    std::cin >> a >> b;

    long long limit = gcd(a, b), count = 1;
    for(long long i = 2; i <= limit; ++i){
        if(isPrime(i) && 0 == limit % i){
            ++count;
        }
    }

    std::cout << count << std::endl;

    return 0;
}