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>
using namespace std;
#define int long long
typedef long long ll;
const int MOD = (int)1e9 + 7;
const int INF = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define REP(i, n) for (int i = 0; i < (n); i++)
#define FOR(i,s,n) for (int i = s; i < (n); i++)
#define ALL(obj) (obj).begin(), (obj).end() //コンテナじゃないと使えない!!
#define debug(x) cerr << #x << ": " << x << "\n";
template <typename T>
ostream& operator<<(ostream& os, vector<T> &V){
    int N = V.size();
    REP(i,N){
        os << V[i];
        if (i!=N-1) os << " ";
    }
    os << "\n";
    return os;
}
int gcd(int a, int b){
    if(b==0) return a;
    return gcd(b,a%b);
}
int solve(int n){
    int copy = n;
    int prime=0;
    for(int i=2;i*i<=copy;i++){
        bool pri = false;
        while(n%i==0){
            n/=i;
            pri=true;
        }
        if(pri){
            prime++;
        }
    }
    if(n!=1) prime++;
    return prime+1; //素因数の数 + 1
}
//fixed<<setprecision(10)<<ans<<endl;
signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int A,B,G;
    cin >> A>>B;
    G=gcd(A,B);
    
    int ans=solve(G);



    cout << ans << endl;
}