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;
#define Would
#define you
#define all(n)  n.begin(),n.end()
#define rall(n) n.rbegin(),n.rend()
typedef long long ll;
const ll INF = 1e18;
const ll MOD = 1e9 + 7;
const double EPS = 1e-10;
const double pi = acos(-1);//3.1415926535897932384626433832795028...
const ll SIZE = 1 << 17;
int dx[] = { 1,0,-1,0 }, dy[] = { 0,1,0,-1 }, alp[30];
ll fac[200005], finv[200005], inv[200005];
vector<ll>dij;
struct edge { ll to, cost; };
vector<vector<edge> >G;

ll mod_pow(ll a, ll b) {
	ll res = 1, mul = a;
	for (int i = 0; i < 31; ++i) {
		if (b >> i & 1) {
			res *= mul;
			res %= MOD;
		}
		mul = (mul * mul) % MOD;
	}
	return res;
}

void addedge(int from, int to, int cost) {
	G[from].push_back({ to,cost });
	G[to].push_back({ from,cost });
}

template<typename T>
vector<T> make_v(size_t a) { return vector<T>(a); }

template<typename T, typename... Ts>
auto make_v(size_t a, Ts... ts) {
	return vector<decltype(make_v<T>(ts...))>(a, make_v<T>(ts...));
}

template<typename T, typename V>
typename enable_if<is_class<T>::value == 0>::type
fill_v(T & t, const V & v) { t = v; }

template<typename T, typename V>
typename enable_if<is_class<T>::value != 0>::type
fill_v(T & t, const V & v) {
	for (auto& e : t) fill_v(e, v);
}

template<typename T>
void outp(vector<T>v) {
	for (int i = 0; i < v.size(); ++i) {
		cout << v[i];
		if (i != v.size() - 1) { cout << " "; }
	}
}

double add(double a, double b) {
	if (abs(a + b) < EPS * (abs(a) + abs(b))) { return 0; }
	return a + b;
}

bool is_prime(int64_t x) {
	for (int64_t i = 2; i * i <= x; i++) {
		if (x % i == 0) return false;
	}
	return true;
}

ll a, b, ans;
int main() {
	cin >> a >> b;
	vector<ll>v;
	for (ll i = 1; i * i <= min(a, b); ++i) {
		if (a % i == 0 && b % i == 0) {
			v.push_back(i);
		}
	}

	for (int i = 0; i < v.size(); ++i) {
		if (is_prime(v[i])) { ++ans; }
	}

	cout << ans << endl;

}