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;

#define rep(i,m,n) for(int i=m;i<n;i++)

long gcd(long a,long b){
 long r;
  while(a%b){
   r=a%b;
    a=b;
    b=r;
  }
 return b;
}

int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);

 long a,b;
  cin >> a >> b;
  
  long lcm=gcd(a,b);
  
  long res=1;
  
  if(lcm%2==0) res++;
  while(lcm%2==0) lcm/=2;
  
  
  for(int i=3;i*i<=lcm;i+=2){
    if(lcm%i) continue;
     res++;
    while(lcm%i==0) lcm/=i;
  } 
  
  if(lcm!=1) res++;
 
  
  
  cout << res << endl;
}