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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
queue <int> sosu;
vector<int> natu(1000001,1);
void magic(){
natu[0]=0;
natu[1]=0;
    for(long  long int i=2; i*i<=1000000000000; i++) {
        if(natu[i]) {
        sosu.push(i);
            for(int j=2; i*j<=1000000000000; j++) {
                natu[i*j] = 0; 
                }
                }
                }
}
int main() {
int64_t a,b;
cin>>a>>b;
int64_t c=__gcd(a,b);
  int count=1;
  magic();
 for(int i=0;i<sosu.size();i++){
   if(c%sosu.front()==0){
     count++;
   }
   sosu.pop();
 }
  cout<<count;
}