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>

// increment
#define rep_n(_1, _2, _3, NAME, ...) NAME
#define rep_2(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep_3(i, initial, n) for (int i = (int)(initial); i < (int)(n); ++i)
#define repeat(...) rep_n(__VA_ARGS__, rep_3, rep_2, rep_1)(__VA_ARGS__)
// decrement
#define rep_r_n(_1, _2, _3, NAME, ...) NAME
#define rep_r_2(i, n) for (int i = (int)(n); i > 0; i--)
#define rep_r_3(i, initial, n) for (int i = (int)(initial); i > (int)(n); i--)
#define repeat_r(...) rep_r_n(__VA_ARGS__, rep_r_3, rep_r_2)(__VA_ARGS__)
// define variable
using namespace std;
typedef long long ll;
using mpint = boost::multiprecision::cpp_int; // 多倍長整数（任意精度）
typedef pair<int, int> P;
const double EPS = numeric_limits<double>::epsilon();
// 最大公約数
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
// 最小公倍数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
// char -> int
int ctoi(char c) { return (int)(c - '0'); }
// 素因数分解
vector< pair<ll, int> > prime_factorize(ll n) {
	ll N = n;
	vector< pair<ll, int> > p;
	for(ll i = 2; i*i <= N; i++) {
		if ((N % i) != 0) continue;
		p.emplace_back(i, 0);
 
		while((N%i) == 0) {
			N /= i;
			p.back().second += 1;
		}
	}
	
	if (n != 1) p.emplace_back(n, 1);
	return p;
}
 
 
int main() {
	ll A, B;
	cin >> A >> B;
 
	ll d = gcd(A, B);
	
	vector< pair<ll, int> > p = prime_factorize(d);
	
	cout << (p.size() + 1) << endl;
	return 0;
}