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 <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < n; i++)
#define repk(i, k, n) for (int i = k; i < n; i++)
#define MOD 1000000007
#define INF 1e9
#define PIE 3.14159265358979323

template <class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
T GCD(T a, T b) {
    if (b == 0)
        return a;
    else
        return GCD(b, a % b);
}
template <class T>
inline T LCM(T a, T b) {
    return (a * b) / GCD(a, b);
}

template <class T>
map<T, T> prime_factorize(T n) {
    map<T, T> res;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            res[i]++;
        }
    }
    if (n != 1)
        res[n] = 1;
    return res;
}

int main() {
    ll a, b;
    cin >> a >> b;
    ll s = GCD(a, b);
    map<ll, ll> m = prime_factorize(s);
    cout << m.size() + 1 << endl;
}