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)

#include <iostream>
#include <iomanip>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
const ll mod=1e9+7;

int main(){
	ll A,B;
	cin >> A >> B;
	ll Max=min(A,B)/3;

	bool flg;
	ll ans=1;
	if(A%2==0 && B%2==0){
		ans++;
	}
	if(A%3==0 && B%3==0){
		ans++;
	}
	vll prime(1,3);
	for(ll i=5; i<=Max; i+=2){
		flg=true;
		for(ll j : prime){
			if(i%j==0){
				flg=false;
				break;
			}
		}
		if(flg){
			if(A%i==0 && B%i==0){
				ans++;
			}
			prime.push_back(i);
		}
	}

	cout << ans << endl;
	return 0;
}