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)

/* 
 Author:N_o_o_B 
 Created:September 28 2019 17:08:09
*/
#include <bits/stdc++.h>
using namespace std;

#define TRACE
#ifdef TRACE
#define trace(...) { cerr<<"[ "; __trace__(#__VA_ARGS__, __VA_ARGS__);}
#undef endl
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const pair<Arg1,Arg2> &x) {
    return out<<"("<<x.first<<","<<x.second<<")";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const vector<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const set<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const map<Arg1,Arg2> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1){
    cerr << name << " : " << arg1 << " ] " << endl;
}
 
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args){
    const string name = names.substr(0,names.find(','));
    cerr<<name<<" : "<<arg1<<" | ";
    __trace__(names.substr(1+(int)name.size()), args...);
}
#else
    #define trace(args...)
#endif
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pair<int,int>> vii;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<pair<ll,ll>> vll;
typedef vector<vl> vvl;
#define fori(i,n) for(int i=0;i<n;i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define ford(i,n)  for(int i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define endl '\n'
clock_t time_p=clock();
void time_taken()
{
    time_p=clock()-time_p;
    cerr<<"Time Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";
}
const ll mod=1e9+7;
const ll INF=1e18;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    cout.precision(12);cout << fixed;

    ll a[2];
    fori(i,2) cin >> a[i];
    set<ll> s[2];
    fori(i,2){
        for(int j=2;j*j<=a[i];j++){
            if(a[i]%j) continue;
            ll cnt=0;
            while(a[i]%j==0){
                a[i]/=j;
                cnt++;
            }
            s[i].insert(j);
        }
        if(a[i]>1) {
          s[i].insert(a[i]);
        }
    }
    set<ll> ans;
    for(ll x:s[0]){
        if(s[1].count(x)) ans.insert(x);
    }
    ll n=ans.size();
    cout << n+1 << endl;
    time_taken();
    return 0;
}