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: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <climits>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <complex>

using namespace std;
using pii = pair<int, int>;
using ll = long long;

ll gcd(ll a, ll b)
{
	return (b == 0) ? a : gcd(b, a % b);
}

bool isp(ll n)
{
	if (n % 2 == 0) return false;
	ll sqn = sqrt(n);
	for (ll i = 3; i <= sqn; i += 2) if (n % i == 0) return false;
	return true;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	vector<int> era(1000001);
	era[1] = era[2] = 1;
	for (int i = 3; i <= 1000000; i += 2) {
		if (era[i] == 0) {
			era[i] = 1;
			for (int j = i + i; j <= 1000000; j += i) era[j] = -1;
		}
	}

	ll a, b;
	cin >> a >> b;
	ll g = gcd(a, b);

	ll sqg = sqrt(g);
	int answer = 0;
	for (ll i = 1; i <= sqg; i++) {
		if (g % i == 0) {
			if (era[i] == 1) answer += 1;
			if (i != g / i && isp(g / i)) answer += 1;
		}
	}
	cout << answer;
	
	return 0;
}