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 <iostream>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<deque>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>

#define rep(i, n) for(int i=0;i<n;i++)

typedef int long long ll;
using namespace std;
typedef pair<int, int> P;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const ll MOD = 1e9 + 7;
static const int MAX = 100;
static const int INF = (1 << 23);

std::vector<bool> IsPrime;

void sieve(size_t max){
    if(max+1 > IsPrime.size()){     // resizeで要素数が減らないように
        IsPrime.resize(max+1,true); // IsPrimeに必要な要素数を確保
    }
    IsPrime[0] = false; // 0は素数ではない
    IsPrime[1] = false; // 1は素数ではない

    for(size_t i=2; i*i<=max; ++i) // 0からsqrt(max)まで調べる
        if(IsPrime[i]) // iが素数ならば
            for(size_t j=2; i*j<=max; ++j) // (max以下の)iの倍数は
                IsPrime[i*j] = false;      // 素数ではない
}
ll gcd(ll a,ll b){
if(a==0||b==0)return max(a,b);
if(b>a)swap(a,b);
 while(1){
        if(a%b!=0){
            ll tmp=b;b=a%b;
            a=tmp;
        }else{
            return b;
        }
    }
}

int main() {
    sieve(1000000);
    ll a,b;
    cin>>a>>b;

    if(b>a)swap(a,b);
    ll ab = gcd(a,b);

    ll ans=0;

    bool f = false;
   // cout<<ab<<endl;

    for(ll i=2;i<=sqrtl(ab);i++){

        if(ab%i==0&&IsPrime[i]){
           // cout<<i<<endl;
            ans++;
        }

        if(i*i!=a){
            ll t=ab/i;

            if(t<=1000000&&IsPrime[t])ans++;
        }
    }



    cout<<ans+1<<endl;

    return 0;
}