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>
using namespace std;

//a<=b 最大公約数
int gcd(int a, int b) {
	if (b % a == 0) {
		return a;
	}
	return gcd(b % a, a);
}
//b=2 素因数分解
vector<int> fa(int a, int b) {
	vector<int> lst;
	while (1) {
		if (b * b > a) {
			lst.push_back(a);
			break;
		}
		if (a % b == 0) {
			lst.push_back(b);
			b = b;
			a = a / b;
			continue;
		}
		b = b++;
	}
	return lst;
}

int main() {
	int a, b;
	cin >> a >> b;
	vector<int> lst = fa(gcd(min(a, b), max(a, b)), 2);
	int in = 2;
	int x = lst.at(0);
	for (int i = 0; i < lst.size(); i++) {
		if (x != lst.at(i)) {
			in = in++;
		}
		x = lst.at(i);
	}
	if (lst.at(0) == 1) {
		cout << 1 << endl;
	}
	else {
		cout << in << endl;
	}
}