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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<unordered_map>
#include<cmath>
#include<map>
using namespace std;
using Graph = vector<vector<int>>;
typedef unsigned long long ll;
ll mod = (10e+8)+ 7;

#define For(N,i) for(int i = 0; i < N; i++)
	
int ma;
	map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int i = 2; i <= sqrt(n); i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
	  ma = max(i,ma);
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
int main(){
	int A,B;
	cin>>A>>B;
	map<int64_t,int> AA  = prime_factor(A);
	map<int64_t,int> AB  = prime_factor(B);
	int cnt = 0;
	for(int i = 0; i <= ma; i++){
		if(AA[i] > 0 && AB[i] > 0){
			cnt++;
		}
	}
	cout<<cnt+1<<endl;
	
	
}