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: Accepted)

import java.util.*;

public class Main{

    public static void main(String args[]){

        Scanner sc = new Scanner(System.in);

		long f = sc.nextLong();
		long s = sc.nextLong();
        long a = Math.min(f, s);
        long b = Math.max(f, s);
        long a_ = (long)Math.ceil(Math.sqrt(a)) + 1;
        long b_ = (long)Math.ceil(Math.sqrt(b)) + 1;
        Set<Long> set_a = new HashSet<Long>();
        Set<Long> set_b = new HashSet<Long>();
        set_a.add(1L);
        set_b.add(1L);

        for(long i = 2; i <= a; i++){
            if(i > a_){
                set_a.add(a);
                break;
            }
            if(a % i == 0) set_a.add(i);
            else continue;
            while(a % i == 0){
                a /= i;
            }
            if(a == 1) break;
        }
        for(long i = 2; i <= b; i++){
            if(i > b_){
                set_b.add(b);
                break;
            }
            if(b % i == 0) set_b.add(i);
            else continue;
            while(b % i == 0){
                b /= i;
            }
            if(b == 1) break;
        }

        long count = 0;
        // if(set_a.size() < set_b.size()){
        //     for(long x: set_a){
        //         if(set_b.contains(x)) count++;
        //     }
        // }else{
        //     for(long x: set_b){
        //         if(set_a.contains(x)) count++;
        //     }
        // }
        for(long x: set_a){
            if(set_b.contains(x)) count++;
        }

        System.out.println(count);

    }

}