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.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long A = sc.nextLong();
		long B = sc.nextLong();
		// ユークリッドの互除法
		long tmp = 0;
		if (A < B) {
			tmp = A;
			A = B;
			B = tmp;
		} // Aが常に大きい

		long r = -1;
		while (r != 0) {
			r = A % B;
			if (r == 0)
				break;
			else {
				A = B;
				B = r;
			}
		} // このときのBが２入力値の最大公約数

		long C = 1;
		Set<Long> memo = new HashSet<Long>();
		long goal = (long)Math.sqrt(B);
		while (Math.sqrt(B) >= C) {
			if (C == 1) {
				memo.add(C);
				C++;
			} else {
				if (B % C == 0) {
					while(B % C == 0) {
						B = B / C;
					}
					memo.add(C);
				}
					C++;
			}
		}
		long ans =0;
		if(B!=1)ans++;
		ans += memo.size();
		System.out.println(ans);
	}
}