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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
bool is_prime[1000011];
void sieve(ll n){
    ll p=0;
    for(ll i=0; i<= n; i++) is_prime[i] = true;
    is_prime[0] = is_prime[1] = false;
    for(ll i=2; i<=n; i++){
        for(ll j = 2*i; j <= n; j+= i)
            is_prime[j] = false;
    }
}

void solve(long long A, long long B){
    //* now A <= B
    ll n = (ll) sqrt((double)A)+1;
    sieve(n);
    vector<ll> pset;
    for(ll p=2; p<=n; p++){
        if(is_prime[p]){
            if(A % p == 0) pset.push_back(p);
        }
    }

    ll cnt = 1;
    for(ll p: pset){
        if(B % p == 0) 
            cnt ++;
    }
    cout << cnt << endl;
}

int main(){
    long long A;
    scanf("%lld",&A);
    long long B;
    scanf("%lld",&B);
    solve(A, B);
    return 0;
}