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>
#define rep(i,a) for(int i=0;i<int(a);++i)
#define REP(i,a,b) for(int i=int(a);i<int(b);++i)
#define pb push_back
#define mp make_pair
#define F first
#define S second
using ll = long long;
using itn = int;
using namespace std;
using Graph = vector<vector<int>>;
static const long double GRATIO=(long double)(1+sqrt(5))/2;
ll GCD(ll a, ll b){
    return b ? GCD(b, a%b) : a;
}
ll LCM(ll a, ll b){
    return a/GCD(a,b)*b;
}
int main(){
    ll A,B;
    cin>>A>>B;
    vector<ll> prime_list;
    vector<ll> table(1000010,0);
    for(ll i=2; i<1e6+5; i++){
        if(table[i]==0){
            prime_list.push_back(i);
            for(ll j=i; j<1e6+5; j+=i){
                table[j]++;
            }
        }
    }
    ll gcd=GCD(A,B);
    ll ans=0;
    for(auto i=prime_list.begin(); i!=prime_list.end(); i++){
        if(gcd%(*i)==0){
            while(gcd%(*i)==0){
                gcd/=(*i);
            }
            ans++;
        }
    }
    if(gcd>1) ans++;
    cout<<ans+1<<endl;
}