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: Time Limit Exceeded)

#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
using namespace std;

ll gcd(ll a, ll b){
    if(b==0) return a;
    else return gcd(b, a%b);
}

bool isPrime(ll n){
    if(n==2) return true;
    for(int i=2; i*i<=n; i++){
        if(n%i==0) return false;
    }
    return true;
}

int main(){
    ll A, B;  cin >> A >> B;
    set<ll> divSet;
    for(int i=1; i*i<=min(A, B); i++){
        if(A%i == 0 && B%i == 0){
            divSet.insert(i);
            ll a = A/i;
            if(A%a == 0 && B%a == 0)
                divSet.insert(a);
            
            ll b = B/i;
            if(A%b == 0 && B%b == 0)
                divSet.insert(b);
        }
    }
    vector<ll> divs;
    for(ll d : divSet){
        if(isPrime(d))
            divs.push_back(d);
    }
    cout << divs.size() << endl;
}