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<bits/stdc++.h>
#define int long long
using namespace std;
const int INF = 1e10;
const int ZERO = 0;

int gcd(int a,int b) {
    if(a == 0 || b == 0) {
        return max(a,b);
    }
    else {
        if(a < b) {
            int tmp = a;
            a = b;
            b = tmp;
        }
        int r = a % b;
        while(r != 0) {
            a = b;
            b = r;
            r = a % b;
        }
        return b;
    }
}

bool IsPrime(int num)
{
    if (num < 2) 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;
}

signed main() {
    int A,B;
    cin >> A >> B;
    int C = gcd(A,B);
    vector<int> vec;
    for(int i = 1;i <= C;i++) {
        if(C % i == 0) {
            vec.push_back(i);
        }
    }
    int ret = 0;
    for(int i = 0;i < vec.size();i++) {
        if(IsPrime(vec.at(i)) == true || vec.at(i) == 1) {
            ret++;
        }
    }
    cout << ret << endl;
}