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 <bits/stdc++.h>
typedef long long ll;
using namespace std;

// 約数列挙
vector<ll> divisor(ll n) {
    vector<ll> ret;
    for(ll i = 1; i * i <= n; i++) {
        if(n % i == 0) {
            ret.push_back(i);
            if(i * i != n) ret.push_back(n / i);
        }
    }
    sort(begin(ret), end(ret));
    return (ret);
}

// 素数か判定
bool isPrimeNumber(ll n) {
    if (n < 2) return false;
    for (ll i = 2; i * i <= n; ++i) {
        if (n % i == 0) return false;
    }
    return true;
}

// 最大公約数を求める
ll gcd(ll x, ll y) {
    if (y == 0) return x;
    return gcd(y, x % y);
}


int main() {
    ll a, b;
    cin >> a >> b;
    ll g = gcd(a, b);
    vector<ll> divisors = divisor(g);
    int ans = 1;
    for (int i : divisors) if (isPrimeNumber(i)) ans++;
    cout << ans << endl;
}