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>
using namespace std;

long long gcd(long long a, long long b) {
	if (a % b == 0) {
		return b;
	}
	else
	{
		return gcd(b, a % b);
	}
}

int main()
{
	long long a, b;
	cin >> a >> b;

	long long c = gcd(a, b);
	long long p = 2;
	long long cnt = 0;

	if (c % p == 0) {
		cnt++;
		while (c % p == 0)
		{
			c /= p;
		}
	}
	p = 3;
	while ((c > 1) && (p * p <= c))
	{
		if (c % p == 0) {
			cnt++;
			while (c % p == 0)
			{
				c /= p;
			}
			p += 2;
		}
	}
	cout << (cnt + 1) << endl;
	return 0;
}