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: Wrong Answer)

#include<bits/stdc++.h>


using namespace std;
int gcd(int a, int b)
{
	if (a < b) {
		a ^= b;
		b ^= a;
		a ^= b;
	}

	return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b)
{
	return a * b / gcd(a, b);
}
bool IsPrime(int num)
{
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

	double sqrtNum = sqrt(num);
	for (int i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			// 素数ではない
			return false;
		}
	}

	// 素数である
	return true;
}

int main()
{
	int n, m;
	int ans = 1;
	cin >> n >> m;
	int g = gcd(n, m);
	set<int> x;
	int num = min(m,n); // 素因数分解する変数num 

	for (int i = 2; i <= g; i++) {
		while (g % i == 0) { // 素数で割り切れなくなるまで割っていく
			if (n % i == 0 && m % i == 0)x.insert(i);
			g /= i;
			//printf("%d %d\n", x[i], num);
		}
	}
	cout << x.size()+1;/*
	//cout << "hoge";
	for(auto &i:x)
	{
		//cout << i << endl;
		if (n % i == 0 && m % i == 0)ans++;
		//ans *= i;//それぞれを+1して掛けていく
	}
	cout << ans << endl;*/
	return 0;
}