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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

function Main(input) {
    input = input.split(" ");        
    input = input.map(e=>parseInt(e,10));

    var maxKouyaku = saiki(input[1],input[0]);    

    var kouyaku = [];

    if(maxKouyaku%2 == 0){
        kouyaku.push(2);
        for(var i=3;i<=maxKouyaku;i+=2){
            if(maxKouyaku%i==0 && !kouyaku.some(e=>i%e==0))kouyaku.push(i);
        }
    }else{
        for(var i=2;i<=maxKouyaku;i++){
            if(maxKouyaku%i==0 && !kouyaku.some(e=>i%e==0))kouyaku.push(i);
        }
    }    
   
    console.log(kouyaku+1);
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));

function saiki(num,shou){    
    if(num%shou==0)return shou;
    return saiki(shou,num%shou)
}