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: Accepted)

#include <bits/stdc++.h>
using namespace std;
vector<long long>v;
vector<long long>vv;
int main()
{
    long long n,m,i,j,k,x,y;
    cin>>n>>m;
    if(n==1 || m==1)
    {
        cout<<1;
        return 0;
    }
    for(long long i=2; i*i<=n; i++)
    {
        if(n%i==0)
        {
            long long temp=n/i;
            v.push_back(i);
            v.push_back(temp);
        }
    }
    for(long long i=2; i*i<=m; i++)
    {
        if(m%i==0)
        {
            long long temp=m/i;
            vv.push_back(i);
            vv.push_back(temp);
        }
    }
    v.push_back(n);
    vv.push_back(m);
    int len=unique(v.begin(),v.end())-v.begin();
    int len2=unique(vv.begin(),vv.end())-vv.begin();
    vector<long long>v1;

    for(i=0; i<len; i++)
    {
        //cout<<v[i]<<endl;
        for(j=0; j<len2; j++)
        {
            if(v[i]==vv[j])
            {
                v1.push_back(v[i]);
                break;
            }
        }
    }
    for(i=0; i<v1.size(); i++)
    {
        //cout<<v1[i]<<endl;
        if(v1[i]!=0)
        {
            for(j=0; j<v1.size(); j++)
            {
                if(v1[j]%v1[i]==0 && i!=j && v1[j]!=0)
                {
                    v1[j]=0;
                }
            }
        }
    }
    x=0;
    for(i=0; i<v1.size(); i++)
    {
        if(v1[i]!=0)
        {
            x++;
        }
    }
    cout<<x+1<<endl;
}