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;

bool isprime[1000000];

void eratos(int n){
    for(int i=0;i<n;i++){
        isprime[i]=true;
    }
    for(int i=0;i<2;i++){
        isprime[i]=false;
    }
    for(int i=0;i<n;i++){
        if(i*i>n)break;
        if(isprime[i]){
            int j=i+i;
            while(j<=n){
                isprime[j]=false;
                j+=i;
            }
        }
    }
}

int main()
{
        vector<int >p;
        eratos(1000000);
        for(int i=0;i<=1000000;i++){
            if(isprime[i]){
                p.push_back(i);
            }
        }
        int a,b;
        cin >> a >> b;
        int s=1;
        for(int i=0;i<p.size();i++){
            if(a%p[i]==0&&b%p[i]==0){
                s++;
                a/=p[i];
                b/=p[i];
            }
        }
        if(a==b&&a!=1)s++;
        cout <<s << endl;
    return 0;
}