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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <math.h>
using namespace std;

bool isPrime(int num)
{
    if (num < 1) return false;
    else if (num <= 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }
    // 素数である
    return true;
}

int main() {
    long A;
    long B;
    cin >> A >> B ;

    // A <= Bにする 
    if (A > B) {
        long tmp = A;
        A = B;
        B = tmp;
    }

    long count = 1; // 1は必ず含まれる
    
    if ( 2 <= A){
        if ( A%2 == 0 && B%2 == 0) count ++;
    }

    if ( 3 <= A){
        for( long i = 3; i <= A/2; i+=2 ){
            if(!isPrime(i)){
                continue;
            }

            if( A%i==0 && B%i==0){
                count++;
            }
        }
    }

    cout << count << endl;

    return 0;
}