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;

typedef long long ll;
#define pii pair<int , int>
#define _FastIO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define MAXX 100005

ll max(ll a , ll b){
    if(a > b)   return a;
    return b;
}

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

ll a , b;
ll used[MAXX];
bool p[MAXX];

int main()
{
    _FastIO;

    cin >> a >> b;
    if(a > b)   swap(a , b);
    vector<ll> v;
    for(ll i = 1; i * i <= a; i++){
        if(a % i == 0 && b % i == 0){
            v.push_back(i);
            int k = a / i;
            if(b % k == 0 && k != i)
                v.push_back(k);
        }

    }
    sort(v.begin() , v.end());
    ll n = v.size();
    for(ll i = 0; i < n - 1; i++){
        for(ll j = i + 1; j < n; j++){
            if(gcd(v[i] , v[j]) > 1){
               used[i]++;
               used[j]++;
            }
        }
    }
    for(ll i = 0; i < n - 1; i++){
        for(ll j = i + 1; j < n; j++){
            if(gcd(v[i] , v[j]) > 1){
               if(!p[i] && !p[j]){
                    if(used[i] > used[j])
                        p[i] = true;
                    else
                        p[j] = true;
               }
            }
        }
    }
    ll ans = 0;
    for(ll i = 0; i < n; i++){
        if(!p[i])
            ans++;
    }
    cout << ans << endl;
    return 0;
}