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;

#define FastRead        ios_base::sync_with_stdio(0);cin.tie(NULL);
#define endl            "\n"
#define NUM             10000005
typedef long long ll;

bool mark[NUM];
map<ll,ll>mp;

void seive()
{
    mark[1]=true;
    for(ll i=4;i<=NUM;i+=2)
        mark[i]=true;

    for(ll i=3;i*i<=NUM;i+=2)
    {
        if(!mark[i])
        {
            for(ll j=i*i;j<=NUM;j+=2*i)
                mark[j]=true;
        }
    }
    return;
}


int main()
{
    seive();
    FastRead

    ll a,b,ans=1;

    cin>>a>>b;

    for(ll i=2;i<=sqrt(a);i++)
    {

        if(!(a%i))
        {
            if(mark[i]==false)
                mp[i]++;
            if(mark[a/i]==false)
                mp[a/i]++;
        }
    }


    for(ll i=2;i<=sqrt(b);i++)
    {

        if(!(b%i))
        {
            if(mark[i]==false)
                mp[i]++;
            if(mark[b/i]==false)
                mp[b/i]++;
        }
    }

    for(auto it:mp)
    {
        if(it.second>=2)
            ans++;
    }

    cout<<ans<<endl;
}