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)

#include <bits/stdc++.h>

using namespace std;

#define intt long long
#define FAST ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

const int N = 100005;

intt a, b;

int main() {

	FAST;

	cin >> a >> b;
	vector <intt> diva, divb;
	intt c = a;
	for (int i = 2; 1LL * i * i <= a; i ++) {
		if (c % i == 0) {
			while (c % i == 0) {
				c /= i;
			}
			diva.push_back(1LL * i);
		}
	}
	if (c != 1) diva.push_back(c);
	c = b;
	for (int i = 2; 1LL * i * i <= b; i ++) {
		if (c % i == 0) {
			while (c % i == 0) {
				c /= i;
			}
			divb.push_back(1LL * i);
		}
	}
	if (c != 1) divb.push_back(c);
	int ans = 0;
	for (int i = 0; i < diva.size(); i ++) {
		if (*lower_bound(divb.begin(), divb.end(), diva[i]) == diva[i]) {
			ans ++;
		}
	}
	ans ++;
	cout << ans << '\n';
	return 0;
}