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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
	long long a, b;
	cin >> a >> b;
	if(b>a)swap(a, b);
	long long r = a % b;
	while(r!=0)
	{
		a = b;
		b = r;
		r = a % b;
	}//bが最大公約数
	
	int num = b;
	vector<int> x(b + 1);
	for(int i=2; i<=b; i++)
	{
		while(num%i == 0)
		{
			x.at(i) = 1;
			num /= i;
		}
	}
	
	int ans = 1;
	for(int i=0; i<=b; i++)
	{
		ans += x.at(i);
	}
	cout << ans << endl;
	return 0;
}