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: Runtime Error)

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _GLIBCXX_DEBUG
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;

long long gcd(long long a, long long b) {
	if (a % b == 0) return b;
	return gcd(b, a % b);
}

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

	int A, B;
	cin >> A >> B;

	long long x;
	x = gcd(A, B);

	vector<long long> ver(0);
	for (long long i = 2; i * i <= x; i++) {
		if(x % i == 0) {
			ver.push_back(i);
			while(x % i == 0) {
				x /= i;
			}
		}
	}

	cout << ver.size() + 2 << "\n";
}