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 <bits/stdc++.h>
#define lli long long int
using namespace std;
lli gcd(lli a, lli b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
} 
bool prime[1000001]; 
void sieve(lli n) 
{ 
    
    memset(prime, true, sizeof(prime)); 
  
    for (lli p=2; p*p<=n; p++) 
    { 
        
        if (prime[p] == true) 
        { 
                for (lli i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
} 
int main()
{
    
    lli a, b;
    cin >> a >> b;
    lli c = gcd(a,b);
    lli count = 1;
    sieve(1000000);
    for(lli i = 2; i <= sqrt(c); i++)
    {
        if(prime[i])
            {
                if(sqrt(c)*sqrt(c) == c && c != 1  && i == sqrt(c))
                {
                    if(prime[(lli)sqrt(c)])
                    count++;
                }
                
                else if(c % i == 0)
                {
                    count++;
                    if(c/i < 1000000 && prime[c/i])
                    {
                        count++;
                    }
                }
            }
    }
    /*if(sqrt(c)*sqrt(c) == c && c != 1)
    {
        if(prime[(lli)sqrt(c)])
        cout << sqrt(c) << " ",count++;
    }*/
    cout << count << endl;
    return 0;
}