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: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <numeric>
#include <math.h>
#include <set>
#include <bitset>
#include <stack>

using namespace std;

typedef long long ll;

const ll prime = 10000000000 + 7;

/*
vector<int> z_algorithm(const string& s) {
	vector<int> prefix(s.size());
	for (int i = 1, j = 0; i < s.size(); i++)
	{
		if (i + prefix[i - j] < j + prefix[j]) {
			prefix[i] = prefix[i - j];
		}
	}
}
*/

int main() {
	
	ll a, b;
	cin >> a >> b;

	// gcd
	ll gcd, c;
	if (a < b) {
		c = a;
		a = b;
		b = c;
	}
	// a >= b
	while (true) {
		c = a % b;
		if (c == 0) {
			gcd = b;
			break;
		}
		else {
			a = b;
			b = c;
		}
	}

	vector<ll> primeList;

	for (ll i = 2; i < pow(gcd, 0.5) + 1; i++)
	{
		if (gcd % i == 0) {
			// i で割れる
			primeList.push_back(i);
			while (true) {
				if (gcd % i == 0) {
					gcd /= i;
				}
				else {
					break;
				}
			}
		}
		if (gcd == 1) {
			break;
		}
		if (i * i > gcd) {
			primeList.push_back(gcd);
			break;
		}
	}

	cout << (1 + primeList.size()) << endl;


	return 0;
}