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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
const ll INF = 1LL<<60;
const ll MOD = 1e9 + 7;

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

long long LCM(long long a, long long b) {
    return (a/GCD(a,b))*b;   //先にGCDで割ることで、overflow防止
}

bool IsPrime(long long 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;
}

int main() {
    ll A,B;
    cin >> A >> B;
    ll G = GCD(A,B);
    vector<ll> vec; //公約数のリスト
    for (ll i = 1;i*i<=G;i++) {
        if (G%i==0) {
            vec.push_back(i);
            vec.push_back(G/i);
        }
    }
    ll cnt = 0;
    for (ll i = 0;i<vec.size();i++) {
        if (IsPrime(vec.at(i))) {
            cnt++;
        }
    }
    cnt++;
    cout << cnt << endl;
}