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 <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include<cmath>
#include<limits>
#define ll long long
#define ALL(x) (x).begin(),(x).end()
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define int_INF 2147483647
#define pint_INF 2000000000
#define ll_INF 	9223372036854775807
#define MOD 100000007
#define vi vector<int>
#define vvi vector<vector<int>>
#define vll vector<long long>
#define pii pair<int,int>
#define vpii vector<pair<int,int>>
using namespace std;

bool is_prime[100000000 + 1];// Up to 1e8 + 1
template<typename T>
vector<T> Eratosthenes(const T N) { // return primes up to N
	vector<T> P;
	for (int i = 0; i <= N; i++)
	{
		is_prime[i] = true;
	}
	for (int i = 2; i <= N; i++)
	{
		if (is_prime[i])
		{
			for (int j = 2 * i; j <= N; j += i)
			{
				is_prime[j] = false;
			}
			P.emplace_back(i);

			//if (P.size() >= N) break;
		}
	}
	return P;
}

template<typename T>
T gcd(T a, T b) {
	if (a > b) {
		T c = a;
		a = b;
		b = c;
	}
	if (a == 0) {
		return b;
	}

	return gcd(b % a, a);
}

int main() {
	ll A, B; cin >> A >> B;
	vector<ll> Erast = Eratosthenes(static_cast<ll>(1000004));
	//cout << *(Erast.end()-1) << endl;
	ll cnt = 1;
	
	for (int i = 0; Erast[i] <= min({ A, B,static_cast<ll>(78500) }); i++) {
		if (A % Erast[i] == 0 or B % Erast[i] == 0) {
			if (A % Erast[i] == 0 and B % Erast[i] == 0) cnt++;
			while (A % Erast[i] == 0) {
				A /= Erast[i];
			}

			while (B % Erast[i] == 0) {
				B /= Erast[i];
			}

		}
	}
	if (A == B and A > 1) {
		cnt++;
	}
	cout << cnt << endl;
}