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,s,N) for (int i=s;i<(N); ++i)
#define ALL(x) (x).begin(), (x).end()
typedef long long ll;
using namespace std;

int ans = 0;

bool is_prime(const ll a){
	if(a<=3){
		return a>1;
	}else if(a%2 == 0 || a%3 ==0){
		return false;
	}else{
		for(ll i=5;i*i<=a;i+=6){
			if(a%i == 0 || a%(i+2) == 0){
				return false;
			}
		}
		return true;
	}
}

int main(){
	ll A,B;
	cin >> A >> B;
	while(B != 0){
      ll r = A%B;
      A = B;
      B = r;
    }
  if(A == 1){
    ans = 0;
  }else if(A<=5){
    ans = 1;
  }else{
    if(A%2==0) ans++;
    if(A%3==0) ans++;
    for(ll i=5; i*i < A+1; i+=6){
        if(A % i == 0 && is_prime(i)) ans++;
        if(A % (i+2) == 0 && is_prime(i+2)) ans++;
    }
  }
	
    cout << ans+1 << endl;
	return 0;
}