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>
#define REP(i, n) for (int (i) = 0; (i) < (int)(n); i++) 
#define RREP(i, n) for(int (i)=((int)(n)-1); (i)>=0; i--)
#define RFOR(i, a, b) for(int (i) =((int)(b)-1); (i)>=(int)a; i--)
#define ALL(v) (v).begin(),(v).end()
#define MOD 1000000007
#define FI first
#define SE second
#define MP make_pair
#define PB push_back
#define SZ(x)  (int)x.size()
#define SP(x)  setprecision((int)x)


using namespace std ;
typedef long long ll;
typedef vector<int> vint;
typedef vector<vint> vvint;
typedef vector<string> vstr;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll LINF = 1e18;
const double EPS = 1e-9;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} //最小公倍数


int main()
{
  ll a, b ;
  cin >> a >> b ;
  ll d = min(a,b) ;
  ll cnt = 0 ;
  ll cntt= 0 ;
  ll c = gcd(a,b) ;
  
  for(ll i=1; i<=c; i++){
    if(i*i>d){
      break ;
    }
    else{
      if(a%i==0 && b%i==0){
        ll e = sqrt(d) ;
        for(ll j=1; j<=d; j++){
          if(i%j==0){
            cntt++ ;
          }
        }
        if(cntt<=2){
          cnt++ ;
        }
      }
      cntt = 0 ;
    }
  }
  
  cout << cnt <<endl ;
    
    
  
  return 0 ;
}