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>
using namespace std;
using pint = pair<int, int>;
using ll = long long;
using ull = unsigned long long;
using pll = pair<ll, ll>;
#define FOR(i, begin, end) for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define VREP(s, ite) for (auto ite = s.begin(); ite != s.end(); ++ite)
#define FI first
#define SE second
#define ALL(v) v.begin(),v.end()
//#define endl "\n"
#define ciosup  \
    cin.tie(0); \
    ios::sync_with_stdio(false);
#define eb emplace_back
#define vint vector<int>
constexpr ll INF = 1e9 + 7;
constexpr ll MOD = 1e9 + 7;

template <typename T>
void vecin(vector<T> &v, int n = 0)
{
    if (n == 0)
        n = v.size();
    for (int i = 0; i < n; ++i)
    {
        cin >> v[i];
    }
}
template <typename T>
void vecout(vector<T> &v, int n = 0)
{
    if (n == 0)
        n = v.size();
    for (int i = 0; i < n - 1; ++i)
    {
        cout << v[i] << " ";
    }
    cout << v[n - 1] << endl;
}

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

int main()
{
    ll a, b;
    cin >> a >> b;
    ll d = gcd(a, b);
    if (d == 1){
        cout << 1 << endl;
        return 0;
    }
    ll ans = 1, now = d;
    int i = 2;
    while (now > 1)
    {
        if (now % i == 0)
        {
            ++ans;
            while (now % i == 0)
            {
                now /= i;
            }
        }
        ++i;
    }
    cout << ans << endl;
}