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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;

namespace
{
	ll GCD(ll a, ll b)
	{
		return a == 0 ? b : GCD(b % a, a);
	}
}

void ABC142D()
{
	ll A, B;
	cin >> A >> B;
	ll C = GCD(A, B);
	ll sqrtC = static_cast<ll>(sqrt(static_cast<long double>(C)));

	vector<ll> x;
	x.push_back(1);
	int i = 2;
	while (i <= C)
	{
		while (C % i == 0)
		{
			if(x.back() != i)
				x.push_back(i);
			C /= i;
		}
		++i;
	}
	int N = x.size();
	cout << N << endl;
}
int main()
{
	ABC142D();
	return 0;
}