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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <iomanip>
#include <queue>
#include <unordered_map>
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
using namespace std;
typedef long long ll;

template<class T>T gcd(T a, T b){if(a<b){a^=b;b^=a;a^=b;}return b?gcd(b,a%b):a;}
template<class T>T lcm(T a, T b){return a*b/gcd(a,b);}
static vector<int> P={2};
template<class T>T next_prime(T x){
    int size=P.size();
    
    if(x<P[size-1]){
        int begin=0,end=size;
        while(end-begin>1){
            int c=(begin+end)/2;
            if(x<P[c]){
                end=c;
            }else{
                begin=c;
            }
        }
        return P[begin];
    }

    T res=P[size-1];
    while(true){
        res++;
        bool prime=true;
        for(int i=0;i<size;i++){
            if(P[i]<=sqrt((double)res)){
                if(res%P[i]==0){
                    prime=false;
                    break;
                }
            }else{
                break;
            }
        }

        if(prime){
            P.push_back(res);
            size++;
            if(res>x){
                break;
            }
        }
    }

    return res;
}

ll A,B;

int main(){
    cin>>A>>B;

    ll g=gcd(A,B);
    ll sg=sqrt(g);
    ll div=1;
    ll ans=1;
    bool b;
    while(div<sg){
        div=next_prime(div);
        if(g%div==0){
            ans++;
            while(g%div==0) g/=div;
        }
    }
    if(g!=1) ans++;

    cout<<ans<<endl;
}