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;
#define rep(i, s, n) for (int i = (s); i < (int)(n); i++)
#define all(v) v.begin(), v.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
using ll = long long;
const ll MOD = 1e9+7; ll LLINF = 1LL << 60; int INF = INT_MAX;

//

int main(){
    ll a,b; cin>>a>>b;
    ll g=__gcd(a,b);
    // gの素因数の個数を求める.
    ll cnt=0;
    queue<ll> que;
    rep(i,2,ceil(sqrt(g))+1){
        if(g%i==0){
            que.push(i);
            que.push(g/i);
        }
    }
    while(!que.empty()){
        ll fnt=que.front();
        if(g%fnt==0){
            cnt++;
            g/=fnt;
        }
        while(g%fnt==0){
            g/=fnt;
        }
        que.pop();
    }
    cout<<cnt+1<<endl;
}