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<iostream>
#include<string>
#include<vector>
#include<math.h>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<stack>
#include<sstream>
#include<iomanip>
#include<map>
#include<unordered_map>
#include<list>
#include<numeric>
typedef long long ll;

#define pi 3.14159265358979323846
#define VI vector<int>
#define VLL vector<long long>
#define MAX max_element
#define MIN min_element
#define all(v) v.begin(), v.end()
const ll MOD = 1e9 + 7;
using namespace std;
template<typename T>
T gcd(T a, T b) {
	if (b == 0) return a;
	return gcd(b, a % b);
	//return b ? gcd(b, a % b) : a;
}
template<typename T>
T lcm(T a, T b) {
	ll g = gcd(a, b);
	return a / g * b; // Be careful not to overflow
	//return a * b / gcd(a, b);
}

template<typename T>
T binarysearch(vector<T> A, T key) {
	ll left = 0;ll right = 4;ll mid;
	while (left < right) {
		mid = (left + right) / 2;
		if (key == A[mid]) return 1;
		if (key > A[mid]) left = mid + 1;
		else if (key < A[mid]) right = mid;
	}
	return 0;
}

template<typename T>
T finder(vector<T> vec, T number) {
	auto itr = find(vec.begin(), vec.end(), number);
	size_t index = distance(vec.begin(), itr);
	if (index != vec.size()) { // 発見できたとき
		return 1;
	}
	else { // 発見できなかったとき
		return 0;
	}
}
ll frac(ll n)//階乗
{
	if (n == 0) {
		return 1;
	}
	return (n * frac(n - 1)) % MOD;
}

template<typename T>
T keta(T a) {
	ll kazu = 1;
	while (1) {
		if (a / 10 != 0) { a /= 10;kazu++; }
		else break;
	}
	return kazu;
}//桁数

//素数判定
template<typename T>
bool IsPrime(T 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;
}

template<typename T>
T nCr(ll n, T r) {
	ll ans = 1;
	for (int i = n; i > n - r; --i) {
		ans = ans * i;
	}
	for (int i = 1; i < r + 1; ++i) {
		ans = ans / i;
	}
	return ans;
}
/*char r[4];
int yaju[4];
char op[3];
int sum = 0;
scanf("%s", r);
for (int i = 0;i < 4;i++) {
	yaju[i] = r[i] - '0';
}
*/
//vector<vector<int>> data(3, vector<int>(4));
//vector<vector<Type>> vv(n);

int main() {
	ll A, B, ans = 1;cin >> A >> B;
	vector<ll>sum;
	vector<vector<ll>> s;
	ll L = gcd(A, B);
	if (IsPrime(L) == 1) {
		cout << 2 << endl;
	}
	else {
		ll i = 1;
		while (1) {
			if (L % i == 0) {
				L /= i;
				if (IsPrime(i)) ans++;
			}
			if (i > L)break;
			i++;
			if (i > 10000000)break;
		}
		cout << ans << endl;
	}
}