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.*;
//素因数分解
class Main{
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        long a = scanner.nextLong();
        long b = scanner.nextLong();
        TreeMap<Long,Long> mapA = primeFactorize(a);
        TreeMap<Long,Long> mapB = primeFactorize(b);
        mapA.put(1L, 1L);
        mapB.put(1L, 1L);

        long cnt =0;
        for(Long key:mapA.keySet()){
 
			if(mapB.containsKey(key)){
				cnt++;
			}
		}
		long ans = cnt;
		System.out.println(ans);
    }

    public static TreeMap<Long, Long> primeFactorize(long N) {
	TreeMap<Long, Long> map = new TreeMap<>();
	for (long i = 2; i * i <= N; i++) {
		long k = 0;
		while (N % i == 0) {
			k++;
			N /= i;
		}
		if(k!=0)
			map.put(i, k);
	}
	map.put(N, 1L);
	return map;
}

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