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.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		long a;
		long b;
		Scanner sc = new Scanner(System.in);
		a = sc.nextLong();
		b = sc.nextLong();
		sc.close();
		long gcd = gcd(a, b);
		Map<Long, Integer>  result = primeFactorize((long)gcd);
		System.out.println(result.size() + 1);
	}

	public static long gcd(long a, long b) {
        long t;
        if (b == 0) return a;
        while (a % b > 0) {
            t = b;
            b = a % b;
            a = t;
        }
        return b;
    }

	public static Map<Long, Integer> primeFactorize(long p) {
        Map<Long, Integer> factor = new HashMap<>();
        if ((p & 1) == 0) {
            int c = 0;
            do {
                c++;
                p >>= 1;
            } while ((p & 1) == 0);
            factor.put(2L, c);
        }
        for (long i = 3; i * i <= p; i += 2) {
            if (p % i == 0) {
                int c = 0;
                do {
                    c++;
                    p /= i;
                } while ((p % i) == 0);
                factor.put(i, c);
            }
        }
        if (p > 1) {
            factor.put(p, 1);
        }
        return factor;
    }
}