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)

/*
	27 January 2020
	islingr
*/

#pragma region Boilerplate
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
 
using ll = long long;
using ld = long double;

using pii = pair<int, int>;
using pll = pair<ll, ll>;

using vi = vector<int>;
using vl = vector<ll>;

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front
#define bk back
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 
#define fi first
#define se second

#define rep(i, a, b) for (auto i = (a); i < (b); ++i) // [a, b)
#define per(i, a, b) for (auto i = (b) - 1; i >= (a); --i) // (b, a]
#define trav(e, x) for (auto &e : x)

template<class T>
bool ckmin(T& a, const T& b) { 
	return a > b ? a = b, 1 : 0;
}
template<class T>
bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0;
}

mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());

template <class T> using Tree = tree<T, null_type, less<T>, 
	rb_tree_tag, tree_order_statistics_node_update>;
#define ook order_of_key
#define fbo find_by_order

namespace helper {
	template<class T, class = decltype(begin(declval<T>()))>
	inline constexpr auto is_iterable(int) { return true_type{}; }

	template<class>
	inline constexpr auto is_iterable(unsigned) { return false_type{}; }
}

namespace input {
	void re(string &s) { cin >> s; }

	template<class T> void re(T& x);
	template<class P, class Q> void re(pair<P, Q>&);
	template<class T, class... Ts> void re(T& t, Ts&... ts);
	
	template<class T> void re(T& x, false_type) { cin >> x; }
	template<class T> void re(T& x, true_type) {
		for (auto &v : x) re(v);
	}
	template<class T> void re(T& x) {
		re(x, helper::is_iterable<T>(0));
	} 
	template<class P, class Q>
	void re(pair<P, Q> &p) {
		re(p.first, p.second);
	}
	template<class T, class... Ts> void re(T& t, Ts&... ts) {
		re(t);
		re(ts...);
	}
}

namespace output {
	template <typename T> void pr(const T&);
	template<class T, class... Ts> void pr(const T&, const Ts&...);

	template<class S, class T>
	void pr(const pair<S, T> &p) {
		pr("(", p.first, ",", p.second, ")");
	}

	template<class T>
	void pr(queue<T> q) {
		pr("{"); bool c = false;
		while (!q.empty())
			pr(c ? ", " : "", q.front()),
			q.pop(),
			c = true;
		pr("}");
	}
	template<class T>
	void pr(priority_queue<T> q) {
		pr("{"); bool c = false;
		while (!q.empty())
			pr(c ? ", " : "", q.top()),
			q.pop(),
			c = true;
		pr("}");
	}
	template<class T>
	void pr(stack<T> s) {
		pr("{"); bool c = false;
		while (!s.empty())
			pr(c ? ", " : "", s.top()),
			s.pop(),
			c = true;
		pr("}");
	}
	void pr(const string &s) { cerr << s; }

	template<typename T>
	void pr(const T& x, false_type) { cerr << x; }

	template<typename T>
	void pr(const T& x, true_type) {
		pr("{"); bool f = false;
		for (const auto &v : x)
			pr(f ? ", " : "", v),
			f = true;
		pr("}");
	}

	template <typename T>
	void pr(const T& x) {
		pr(x, helper::is_iterable<T>(0));
	}
	
	template<class T, class... Ts>
	void pr(const T& t, const Ts&... ts) { pr(t); pr(ts...); }
	
	void pc() { pr("]\n"); } // debug w/ commas

	template<class T, class... Ts>
	void pc(const T& t, const Ts&... ts) { 
		pr(t);
		if (sizeof...(ts)) pr(", ");
		pc(ts...); 
	}

	#define dbg(x...) pr("[", #x, "] = ["), pc(x);
}

struct IO {
	IO(bool b) {
		if (b) fastio();
		cerr << fixed << setprecision(numeric_limits<ld>::digits10);
	}
	IO(bool b, string s) : IO(b) {
		setIn(s + ".in");
		setOut(s + ".out");
	}
	void fastio() {
		ios_base::sync_with_stdio(false);
		cin.tie(nullptr);
	}
	void setIn(string s) { freopen(s.c_str(), "r", stdin); }
	void setOut(string s) { freopen(s.c_str(), "w", stdout); }
};
#pragma endregion 

using namespace input;
using namespace output;
IO io = IO(true);

bool is_prime(ll p) {
	for (ll d = 2; d * d <= p; ++d)
		if (p % d == 0)
			return false;
	return true;
}

ll solve() {
	ll a, b; re(a, b);
	ll g = __gcd(a, b);
	int ans = 1;
	if (g % 2 == 0) ++ans;
	while (g % 2 == 0) g /= 2;
	for (ll d = 3; d * d <= g; d += 2) {
		if (g % d == 0) {
			while (g % d == 0) g /= d;
			++ans;
		}
	}
	if (g > 2) ++ans;
	return ans;
}

signed main() {
	pr(solve());
}