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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cmath>
using namespace std;

bool prime[1000005];

long long gcd (long long A, long long B){
    while (A != 0){
        long long C = A;
        A = B % A;
        B = C;
    }
    return B;
}

void setup(long long N){
    fill(prime, prime+N, true);
    for (long long i=2; i<=N; i++){
        if (prime[i]){
            for (long long j=2*i; j<=N; j+=i){
                prime[j] = false;
            }
        }
    }
}

int main() {
    long long A, B;
    cin >> A >> B;
    long long c = gcd(A, B);
    setup((int) sqrt(c) + 5);
    long long tot = 0;

    bool works = true;
    for (long long i=2; i<= (long long) sqrt(c) + 5; i++){
        if (prime[i] && c % i == 0) {
            tot++;
            works = false;
        }
    }
    tot++;
    if (works && c != 1) tot++;
    cout << tot << endl;
    return 0;
}