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>
#include <boost/multiprecision/cpp_int.hpp>

/***** type *****/
using namespace std;
using ll = long long;
using ld = long double;
template <class T> using vt = vector<T>;
template <class T> using vvt = vector<vector<T>>;
template <class T> using vvvt = vector<vector<vector<T>>>;
template <class T1, class T2> using umap = unordered_map<T1, T2>;
using ml = boost::multiprecision::cpp_int;

/***** define *****/
#define all(c) (c).begin(), (c).end()            // begin to end
#define rep(i, b, e) for (ll i = b; i < e; i++)  // repeat
#define repr(i, b, e) for (ll i = b; e < i; i--) // repeat reverse
#define val(itr) *(itr)                          // get value
/***** const value *****/
#define llong_max 9223372036854775807            // 9 * 10^18
#define ldbl_max 1.79769e+308                    // 1.7 * 10^308
#define mod 1000000007                           // 1e9 + 7
#define pi 3.14159265                            // 3.14 ...
/***** for each macro *****/
#define fori(i, ...) if(ll i = -1) for(__VA_ARGS__) if(i++, true)
#define each(i, e, c) fori(i, auto e = c.begin(); e != c.end(); ++e)
#define forir(i, v, ...) if(ll i = Size(v)) for(__VA_ARGS__) if(i--, true)
#define eachr(i, e, c) forir(i, auto e = c.rbegin(); e != c.rend(); ++e)

/***** lambda *****/
auto Count = [] // long long count value
(auto b, auto e, auto x) { return (ll)count(b, e, x); };
auto CtoN = [] // char to number
(auto c) { return (ll)(c - '0'); };
auto DivC = [] // long double div ceiling
(auto a, auto b) { return ceil((ld)a / (ld)b); };
auto Fix = [] // fix value
(auto b, auto e, auto fix)
{ for (auto it = b; it != e; ++it)* it += fix; };
auto Pow = [] // long long pow
(auto a, auto b) { return (ll)pow(a, b); };
auto Pow2 = [] // long long pow2
(auto n) { return (1LL << n); };
auto Pow10 = [] // long long pow10
(auto n) { return (ll)pow(10, n); };
auto Size = []  // long long collection size
(auto& c) { return (ll)(c).size(); };
auto Sum = [] // long long accumulate
(auto b, auto e) { return accumulate(b, e, 0LL); };

/***** template *****/
template <class T> vvt<T> VVT
(ll ys, ll xs, T fill = T())
{	// vector<vector<T>> resize + fill
	vvt<T> v(ys);
	each(i, y, v) { val(y).resize(xs, fill); } return v;
}
template <class T> vvvt<T> VVVT
(ll zs, ll ys, ll xs, T fill = T())
{	// vector<vector<vector<T>>> resize + fill
	vvvt<T> v(zs);
	each(i, z, v) { val(z) = VVT(ys, xs, fill); } return v;
}
template <class T> vt<T> InputVT
(ll size, T fix = T())
{	// input vector<T> (T != struct) + fix
	vt<T> v(size);
	each(i, e, v) { cin >> val(e); val(e) += fix; } return v;
}
template <class T> vvt<T> InputVVT
(ll ys, ll xs, T fix = T())
{	// input vector<vector<T>> (T != struct) + fix
	vvt<T> v(ys); each(i, y, v) val(y).resize(xs);
	each(i, y, v) each(j, x, val(y)) { cin >> val(x); val(x) += fix; } return v;
}
template <class T> vvvt<T> InputVVVT
(ll zs, ll ys, ll xs, T fix = T())
{	// input vector<vector<vector<T>>> (T != struct) + fix
	vvvt<T> v(zs);
	each(i, z, v) { val(z) = InputVVT<T>(ys, xs, fix); } return v;
}

namespace NyaGadget
{	// 素数テーブルライブラリ
	void PrimeTable(ll n, vector<ll>& res)
	{
		vector<bool> test(n + 1, true);
		test[0] = test[1] = false;
		for (ll i = 2; i * i <= n; i++)
		{
			if (!test[i]) continue;
			for (ll j = 2 * i; j <= n; j += i)
				test[j] = false;
		}

		res.clear();
		for (ll i = 0; i < Size(test); i++)
			if (test[i]) res.push_back(i);
	}
}

using namespace NyaGadget;

int main(void)
{
	ll A, B; cin >> A >> B;

	vt<bool> adiv(Pow10(6), false);
	rep(i, 1, sqrt(A)) if (A % i == 0) adiv[i] = true;
	vt<bool> bdiv(Pow10(6), false);
	rep(i, 1, sqrt(B)) if (B % i == 0) bdiv[i] = true;

	vt<ll> prime;
	PrimeTable(Pow10(6), prime);
	ll count = 0;
	each(i, e, prime)
	{
		if (adiv[val(e)] && bdiv[val(e)]) count++;
	}

	cout << count + 1;
	return 0;
}