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>
using namespace std;
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
typedef long long ll;
typedef pair<ll,ll> P;
using vi=vector<ll>;
using vvi=vector<vector<ll>>;
using vc=vector<char>;
using vvc=vector<vector<char>>;
const int inf=1000000007;
const int mod=1000000007;
void chmin(auto &a,auto b){if(b<a)a=b;}
void chmax(auto &a,auto b){if(a<b)a=b;}
ll gcd(ll a,ll b){
  while(b!=0){
    if(a>=b)swap(a,b);
    b=b%a;
  }
  return a;
}
int main(){
  int N=5000000;
  ll a,b;cin>>a>>b;
  ll ans=1,n=gcd(a,b),k=2;
  vi arr(N);
  for(int i = 0; i < N; i++){
		arr[i] = 1;
	}
	for(int i = 2; i < sqrt(N); i++){
		if(arr[i]){
			for(int j = 0; i * (j + 2) < N; j++){
				arr[i *(j + 2)] = 0;
			}
		}
	}

	for(int i = 2; i < N; i++){
		if(arr[i]){
			if(n%i==0){
              ans++;
              n/=i;
            }
		}
	}
  if(n>5000000)ans+=2;
  cout<<ans<<endl;
}