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 <cmath>
#include <map>
using namespace std;
using ll = long long;

ll gcd(ll a,ll b){
  if(b == 0) return a;
  return gcd(b,a%b);
}

bool isPrime(ll n){
  if(n == 1) return false;
  if(n == 2) return true;
  if(n%2 == 0) return false;

  for(int i=3;i<=sqrt(n);i++){
    if(n%i == 0) return false;
  }
  return true;
}

map<ll,int> prime_factors(ll n){
  map<ll,int> res;
  if(n == 1){
    res[n] = 1;
    return res;
  }
  ll _n = n;
  for(int i=2;i*i<=_n;i++){
    while(n%i==0){
      res[i]++;
      n /= i;
    }
  }
  if(n != 1) res[n] = 1;
  return res;
}

int main(){
  ll a,b;
  cin >> a >> b;

  ll g = gcd(a,b);

  int ans = 0;
  if(isPrime(g)){
    ans = 2;
  }else{
    map<ll,int> res = prime_factors(g);
    res[1] = 1;
    for(auto itr = res.begin();itr != res.end();itr++){
      ans++;
    }
  }

  cout << ans << endl;
}