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.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long A = Long.parseLong(sc.next());
		long B = Long.parseLong(sc.next());
		if (gcd(A, B)==1) {
			System.out.println(1);
		} else {
			System.out.println(factorization(gcd(A, B)).size()+1);
		}
	}
	static long gcd(long a,long b) {
	    long temp;
	    while ((temp=a%b)!=0) {
	        a=b;
	        b=temp;
	    }
	    return b;
	}
	static ArrayList<long[]> factorization (long n) {
	    ArrayList<long[]> list = new ArrayList<long[]>();
	    long tmp= n ;
	    for (int i=2;i<=Math.sqrt(n);i++) {
	        if (tmp%i==0) {
	            int cnt=0;
	            while (tmp%i==0) {
	                cnt++;
	                tmp/=i;
	            }
	            long[] x = {i, cnt};
	            list.add(x);
	        }
	    }
	    if (tmp!=1) {
	        long[] x = {tmp, 1};
	        list.add(x);
	    }

	    if (list.size()==0) {
	        long[] x = {n, 1};
	        list.add(x);
	    }
	    return list;
	}
}