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.*;

public class Main {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        long A = sc.nextLong();
        long B = sc.nextLong();

        System.out.println(solve(A, B));
    }

    /**
     * https://atcoder.jp/contests/abc142/submissions/7746549
     * AとBで共通する素数の約数の個数+1が解
     * @param A
     * @param B
     * @return
     */
    private static int solve(long A, long B) {
        int count = 1;
        List<Pair> primeFactorA = primeFactorize(A);

        for (Pair p : primeFactorA) {
            // System.err.println(p.first + " ** " + p.second);
            if (B%p.first == 0) {
                count++;
            }
        }

        return count;
    }

    public static class Pair {
        long first;
        long second;

        public Pair(long a, long b) {
            this.first = a;
            this.second = b;
        }
    }

    private static List<Pair> primeFactorize(Long N) {
        List<Pair> list = new ArrayList();

        for (long p=2; p <= N; p++) {
            if (N%p != 0) continue;
            int num = 0;
            while (N%p == 0) {
                num++;
                N /= p;
            }

            list.add(new Pair(p, num));
        }

        return list;
    }
}