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.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.Scanner;
import java.util.HashSet;
import java.util.Iterator;

class Main {

    static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long A = sc.nextLong();
        long B = sc.nextLong();

        HashSet<Long> hsA=primeFactor(A);
        HashSet<Long> hsB=primeFactor(B);

        hsA.add(1l);
        hsB.add(1l);

        HashSet<Long> ans = new HashSet<Long>();

        Iterator ite = hsA.iterator();
        while(ite.hasNext()){
            long num = (long)ite.next();
            if(hsB.contains(num)){
                ans.add(num);
            }
        }

        System.out.println(ans.size());
    }

    public static final HashSet<Long> primeFactor(long n) {
		final HashSet<Long> list = new HashSet<Long>();
		if (n < 2) { 
			return list;
		}

		while (n != 1 && n % 2 == 0) { 
			list.add(2l);
			n >>>= 1;
		}

		for (long i = 3; i * i <= n; i += 2) {
			while (n % i == 0) {
				list.add(i);
				n /= i;
			}
		}

		if (n != 1) {
			list.add(n);
		}
		return list;
	}

}