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 <vector>
#include<queue>
#include<map>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<bitset>
using namespace std;
using vi = vector<int>;
using vii = vector<vi>;
using vs = vector<string>;
using pii = pair<int, int>;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,m,n) for(int i=m;i<n;i++)
#define vsort(v) sort(v.begin(), v.end())
#define rv(v) reverse(v.begin(), v.end())
#define ll long long
#define PI 3.141592653589793
using vll = vector<ll>;
using Graph = vii;
const ll MOD = 1e9 + 7;

int main() {
	ll a, b;
	cin >> a >> b;
	if (a >= b) {
		ll temp = a;
		a = b;
		b = temp;
	}
	vll v(1, 1);
	if (a % 2 == 0 && b % 2 == 0) {
		v.push_back(2);
		a /= 2;
	}
	for (int i = 3; i <= a; i += 2) {
		if (a % i == 0 && b % i == 0) {
			v.push_back(i);
			a /= i;
		}
	}
	int n = v.size();
	cout << n << endl;
}