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: Accepted)

#include<cstdio>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>

using namespace std;

long long gcd(long long a, long long b)
{
	long long temp;

	while (b != 0)
	{
		a %= b;

		temp = a;
		a = b;
		b = temp;
	}

	return a;
}

int main()
{
	long long a, b;
	vector<long long> prime;
	long long i, j;
	bool isprime;
	long long temp;
	long long cnt = 1;
	long long ub;

	scanf("%lld %lld", &a, &b);

	prime.push_back(2);

	temp = gcd(a, b);

	ub = (long long)sqrt(double(temp));

	for (i = 3; i <= ub; i += 2)
	{
		isprime = true;

		for (j = 0; j < prime.size() && prime[j] * prime[j] <= i; j++)
		{
			if (!(i%prime[j]))
			{
				isprime = false;

				break;
			}
		}

		if (isprime)
		{
			prime.push_back(i);
		}
	}

	for (i = 0; i < prime.size(); i++)
	{
		if (temp == 1 || prime[i] * prime[i] > temp)
		{
			break;
		}

		if (!(temp%prime[i]))
		{
			cnt++;

			while (!(temp%prime[i]))
			{
				temp /= prime[i];
			}
		}

		if (temp == 1 || prime[i] * prime[i] > temp)
		{
			break;
		}
	}

	if (temp != 1)
	{
		cnt++;
	}

	printf("%lld\n", cnt);

	return 0;
}