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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) throws Exception {
        
        Scanner sc = new Scanner(System.in);
        long a = sc.nextLong();
        long b = sc.nextLong();
        long c = kys(a,b);
        long res = 1;
        res = res + bk(c,1,0);
        System.out.println(res);
    }
    static long kys(long a,long b){
        if(b == 0){return a;}
        return kys(b,a % b);
    }
    static long bk(long c,long j,long plus){
        if(c == 1){return plus;}else{
            for(long i = j+1;i<=c;i++){
                if(c % i == 0){
                    plus++;
                    return wr(c,i,plus);
                }
            }
        }
        return plus;
    }
    static long wr(long c,long i,long plus){
        while(c % i == 0){
            c = c / i;
        }
        return bk(c,i,plus);
    }
}