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.*;
import java.util.ArrayDeque;
import java.util.Queue;
import java.math.RoundingMode;
import java.math.BigDecimal;


public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		long a = sc.nextLong();
		long b = sc.nextLong();
		long gcd = gcd(a,b);
		long tmp;
		long sum = 0;
		if(gcd == 1) {
			sum++;
		}
		else {
			sum++;
			long N = gcd;
			if(is_prime(N)) {
				sum++;
			}
			else {
			for(long i = 2; i <= N; i++) {
					if(N%i==0) {
						//System.out.println(i);
						sum++;
					}
					while(N%i==0) {
						N = N/i;
						System.out.println(N);
					}
				}
			}
		}
		 System.out.println(sum);
	}
	private static long gcd(long x, long y) {
		long tmp;
		if(x==y){
			return 1;
		}
		else {
			while ((tmp = x % y) != 0) {
				x = y;
				y = tmp;
			}
			return y;
		}
	}
	private static boolean is_prime(long num) {
		if (num < 2) return false;
	    else if (num == 2) return true;
	    else if (num % 2 == 0) return false; 
		double sqrtNum = Math.sqrt(num);
	    for (int i = 3; i <= sqrtNum; i += 2)
	    {
	        if (num % i == 0)
	        {
	            return false;
	        }
	    }
	    return true;
	}
}