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: Time Limit Exceeded)

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    //公約数計算
    while(a>0&&b>0){
        if(a<b){
            b=b%a;
        }
        else{
            a=a%b;
        }
    }
    if(b>0){a=b;}
    //素因数分解
    //素数列の生成
    vector<bool> prime;
    int i=1;
    while (i*i<a){i++;}
    prime.resize(i+1,true);
    prime[1]=false;
    for(int j=2;j<=i;j++){
        if(prime[j]){
            for(int k=2;j*k<=i;k++){
                prime[j*k]=false;
            }
        }
    }
    /*for(int j=0;j<prime.size();j++){
        cout<<prime[j]<<endl;
    }*/
    //素数列に従って調査
    int cnt=0;
    int d=2;
    while(d*d<=a){
        if(prime[d]){
            if(a%d==0){
                cnt++;
                while(a%d==0){
                    a=a/d;
                }
            }
        }
    }
    if(a!=1){cnt=cnt+1;}
    cout<<cnt+1<<endl;
    return 0;
}