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)

#pragma region
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <map>
#include <queue> 
#include <stack>
#include <set>
#include <list>

using namespace std;
typedef long long ll;
//#define rep(i, s, e) for (int(i) = (s); (i) < (e); ++(i))
#define rep(i, e) for (int(i) = 0; (i) < (e); ++(i))
#define all(x) x.begin(),x.end()
#pragma endregion


int main()
{
	ll a, b; cin >> a >> b;
	vector<bool> isPrime(sqrt(max(a, b)) + 10, true);
	isPrime[0] = false;
	for (int i = 2; i < sqrt(max(a, b)) + 5; ++i)
	{
		if (!isPrime[i])continue;

		for (int j = i + i; j < sqrt(max(a, b)) + 5; j += i)
			isPrime[j] = false;
	}
	int res = 0;
	rep(i, sqrt(max(a, b)) + 5)
	{
		ll x = i;
		if (isPrime[x] && a%x == 0 && b%x == 0)++res;
	}
	cout << res << endl;
}