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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;
#define N 100009
#define LL long long
#define INF 0x7fffffff
#define mymin(x,y) (x>=y?y:x)
#define mymax(x,y) (x>=y?x:y)
LL gcd(LL a, LL b) {
	if (!a) return b;
	if (!b) return a;
	if (a > b) return gcd(a % b, b);
	return gcd(a, b % a);
}
int main() {
	std::ios::sync_with_stdio(0); std::cin.tie(0);
	LL a, b;
	cin >> a >> b;
	LL n = gcd(a, b);
	if (n == 1) {
		cout << 1;
		return 0;
	}
	LL p = 2;
	LL ans = 0;
	while (p < 1e6 && n != 1) {
		if (!(n % p)) {
			ans++;
			while (!(n % p)) {
				n /= p;
			}
		}
		p++;
	}
	if (!ans) {
		cout << 2;
		return 0;
	}
	if (n > 1e6) ans++;
	cout << ans + 1;
	return 0;
}