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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <string>
#include <cstring>
#include <map>

using namespace std;
typedef long long ll;

int gcd(ll x, ll y) {
	return y ? gcd(y, x%y) : x;  /*yが0でなければgcd、1であればxをかえす*/
}

int main(){
	ll a, b;
	ll count = 0;
	ll ans = 0;
	cin >> a >> b;
	ll x= gcd(a, b);

	ll c = 2;
	while (x >= c*c){
		if (x % c == 0) {	
			while (x%c == 0){
				x = x / c;
			}
			ans++;
		}
		c++;
	}
	
	cout << ans+2 << endl;
	return 0;
}