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> 
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define _GLIBCXX_DEBUG
typedef long long ll;
using namespace std;
ll gcd(ll n, ll m){
    if(n % m == 0) return m;
    return gcd(m, n % m);
}
int main(){
  ll a,b;
  cin>>a>>b;
  ll l=1;
  ll c=gcd(a,b);
  ll d=c;
  bool ans=true;
  if(c%2==0){l++;
             while(c%2==0){c/=2;}}
  for(ll i=3;i*i<=d;i+=2){
    if(c%i==0){l++;
               ans=false;
               while(c%i==0){c/=i;}}}
  if(ans&&c!=1){l++;}
  cout<<l<<endl;
}