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>

using namespace std;

//FILE *fi=freopen("1.txt","r",stdin);

long long a,b,dem=1;
bool f[1000001];
vector<long long> v;

void sang()
{
    for (int i=2;i*i<=1000000;++i)
    {
        if (!f[i])
        {
            v.push_back(i);
            for (int j=i*2;j<=1000000;j+=i)
                f[j]=1;
        }
    }
}

void process()
{
    sang();
    long long k=__gcd(a,b);
    for (auto i:v)
    {
        if (k%i==0) dem++;
        while (k%i==0)
            k/=i;
    }
    dem+=(k!=1);
    cout<<dem;
}

int main()
{
    scanf("%lld %lld",&a,&b);
    process();
    return 0;
}