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: Wrong Answer)

//package BeginnerContest142;

import java.util.*;
import java.io.*;

public class Main {
    static long MODULO = (long)1e9+7;

    public static long factors(long n, long m){ // returns a list of the prime factors of the number n
        HashSet<Long> fact = new HashSet<>();
        fact.add(1L);
        for(long i = 2; i*i <= n; i++){
            while(n % i == 0){
                if(m % i == 0)
                    fact.add(i);
                n/=i;
            }
        }
        for(long i = 2; i*i <= m; i++){
            while(m % i == 0){
                if(n % i == 0)
                    fact.add(i);
                m/=i;
            }
        }
        //System.out.println(fact);
        return fact.size();
    }
    // can also use sieve of eratosthenes for precomputing prime

    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        PrintWriter pw = new PrintWriter(System.out);
        long a = sc.nextLong();
        long b = sc.nextLong();
        pw.println(factors(a,b));
        pw.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}