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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <chrono>
#include <random>
#include <functional>

using namespace std;
using ll = long long;

#define rep(i, n) for (int i = 0; i < (n); ++i)
#define PI 3.14159265359

const int INF = 1e9 + 7;

const int mod = 1000000007;


ll gcd(ll x, ll y) { return y ? gcd(y,x%y) : x;}

// 素因数分解
// 素因数と何乗かを返す
vector<pair<ll,int>> factorize(ll n) {
  vector<pair<ll,int>> res;
  for(int i=2; i*i<=n; i++){
    if(n%i) continue;
    res.emplace_back(i, 0);

    while(n%i == 0){
      n /= i;
      res.back().second++;
    }
  }

  // √n以上の素因数は高々1個しか現れない
  if(n != 1) res.emplace_back(n,1);

  return res;
}

int main(){
  ll A, B;
  cin >> A >> B;
  ll g = gcd(A, B);

  auto f = factorize(g);

  int ans = f.size() + 1;

  cout << ans << endl;

  return 0;
}