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: Runtime Error)

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

bool IsPrime(int num) {
    if(num <2) return true;
    else if(num == 2) return true;
    else if(num % 2 == 0) return false;

    double sqrtNum = sqrt(num);
    for(int i=3; i<=sqrtNum; i+=2) {
        if(num % i == 0) {
            return false;
        }
    }
    return true;
}

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

int main() {
    int num1, num2;
    cin >> num1 >> num2;

    int g = gcd(num1, num2);
    vector<int> gp;
    gp.push_back(1);
    if(g>1) {
        gp.push_back(2);
        for (int i = 3; i <= g; i += 2){
            if(IsPrime(i)) {
                gp.push_back(i);
            }
        }
    }

    int ans = 0;
    for(auto itr = gp.begin(); itr != gp.end(); itr++) {
        if(num1 % *itr == 0 && num2 % *itr ==0) {
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}