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<vector>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<queue>
#include<math.h>
#define REP(i,a) for (int i = 0;i < (a);++i)
#define FOR(i,a,b) for (int i = (a);i < (b); ++i)
#define FORR(i,a,b) for (int i = (a);i >= (b); --i)
#define ALL(obj) (obj).begin,(obj).end()
#define SORT(list) sort(ALL((list)));
#define MOD 1000000007
using namespace std;
using ll = long long;
int main(){
    ll a,b;
    cin >> a >> b;
    ll A = max(a,b);
    ll B = min(a,b);
    map <int,int>m;
    ll answer = 0;
    bool flag;
    FOR(i,2,ceil(sqrt(A))+1){
        flag = true;
        while(A%i==0){
            A/=i;
            if(B%i==0){
                if(flag){
                    answer++;
                    //cout << i << endl;
                }
                
                flag = false;
            }    
        }
    }
    cout << answer+1 << endl;

    return 0;
}