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 <bits/stdc++.h>
using namespace std;

bool IsPrime(int num){
  if (num < 2) return false;
  else if (num == 2) return true;
  else if (num % 2 == 0) return false; // 偶数はあらかじめ除く
  double sqrtNum = sqrt(num);
  for (int i = 3; i <= sqrtNum; i += 2){
    if (num % i == 0){
      return false;
    }
  }
  return true;
}

int main(){
  long long A, B;
  cin >> A >> B;
  long long t = max(A, B), s = min(A, B);
  while(t % s != 0){
    long long a = t % s;
    t = s;
    s = a;
  }
  queue<long long> q;
  for(long long i= 1; i <= s; i++){
    if(s % i ==0){
      q.push(i);
    }
  }
  long long ans = 1;
  while(!q.empty()){
    long long p = q.front();
    if(IsPrime(p)){
      ans++;
    }
    q.pop();
  }
  cout << ans << endl;
}