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>
#include<iomanip>
using namespace std;
typedef long long ll;
typedef pair<int,int> pl;

ll gcd(ll a,ll b){
  if(a<b)swap(a,b);
  ll c=a%b;
  while(c!=0){
    a=b;
    b=c;
    c=a%b;
  }
  return b;
}

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)
    {
      // 素数ではない
      return false;
    }
  }

  // 素数である
  return true;
}

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


  ll c=gcd(a,b);
  bool z=IsPrime(c);
  ll num=c;
  int cnt=0;
  if(z)
  cnt++;
  else{
    for(ll i=2; i<=c; i++){
      bool chk=1;
      while(num%i==0){
        if(chk){
          cnt++;
          // cout<<"i "<<i<<endl;
        }
        num/=i;
        chk=0;
      }
      if(num==1)
      break;
    }
  }
  // cout<<"cnt "<<cnt<<endl;
  // cnt=(cnt+1)*cnt/2;
  cout<<cnt+1<<endl;


}