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>
#define ll long long int
using namespace std;
int gcd(ll a, ll b)
{
		if(a==0)
		return b;
		
		if(b==0)
		return a;
		
		if(a<b)
		return gcd(a,b-a);
		
		if(a>b)
		return gcd(a-b,b);
	}
int main()
{
	
	ll a,b;
	cin>>a>>b;
	int x = gcd(a, b);
	int count;
	
	ll max = (a>b ? a:b);
	bool prime[max];
	for(int i =0; i<max;i++)
	{
		prime[i]=true;
	}
	prime[0]=prime[1]= false;
	
	for (int p = 2; p * p <= max; p++) { 
  
        // If prime[p] is not changed, then it is a prime 
        if (prime[p] == true) { 
  
            // Update all multiples of p as non-prime 
            for (ll i = p * p; i <= max; i += p) 
                prime[i] = false; 
        } 
    } 
    
for (int i = 2; i <= x; i++) { 
  
        // If i is prime and a divisor of gcd 
        if (prime[i] && x % i == 0) 
		{ 
            count++;
        } 
    } 
	
	cout<<count+1;
	return 0;
}