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)

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

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

			long a = sc.nextLong();
			long b = sc.nextLong();
			
			long gcd = gcd(a, b);

			Map<Long, Integer> factors = new HashMap<>();

			// iterate 2 to GCD
			List<Long> numberList = new ArrayList<>();
			for(long i = 2 ; i <= gcd ; i++ ) {
				numberList.add(i);
			}
			
			while(numberList.size() > 0 ) {
				long p = numberList.remove(0);
				
				if ( gcd % p == 0 ) {
					factors.put(p, 1);
					//remove from numberList
//					for(int j = 0 ; j < numberList.size(); j++ ) {
//						long val = numberList.get(j);
//						if ( val % p == 0 ) {
//							numberList.remove(val);
//						}
//					}
					
					long val = p;
					while ( val <= gcd ) {
						numberList.remove(val);
						val = val * 2;
					}
				}
			}

			System.out.println(factors.size() + 1);
		}
	}

	public static long gcd(long a, long b) {
		if (a == 1 || b == 1 ) return 1;
		if ( a == b ) return a;
		if ( a > b) {
			return gcd(a-b, b);
		}
		else {
			return gcd(a, b-a);
		}
	}

}