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>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

vector<long long int> pri;

bool isprime(long long int x){

	if(x%2==0 && x!=2) return false;

	int sqrtn;
	sqrtn=(int)sqrt(x)+1;

	for(int i=0;i<pri.size();i++){
		if(x%pri[i]==0) return false;
		if(sqrtn<pri[i]) return true;
	}
	return true;
}

//n以下の素数を配列に
void list_prime(int n){
	int flag, sqrtn;
	pri.emplace_back(2);

	for(long long int i=3;i<=n;i+=2){
		flag=0;
		sqrtn=(int)sqrt(n)+1;
		for(int j=0;j<pri.size();j++){
			if(i%pri[j]==0){
				flag=1;
				break;
			}
			if(sqrtn<pri[j]) break;
		}
		if(flag==0) pri.emplace_back(i);
	}
}

//b-a>1000000
//b>a

int main(){

	long long int a, b, ans=1, lim;

	cin>> a >> b;

	lim=sqrt(max(a, b))+1;
	list_prime((int)lim);

	if(isprime(a)) return 0;

	for(int i=0;i<pri.size();i++){
		if(a%pri[i]==0 && b%pri[i]==0){
			ans++;
			//cout<< pri[i] <<endl;
		}
	}

	cout<< ans <<endl;

}