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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <time.h>
#include <math.h>
#include <map>
#include <algorithm>

using namespace std;

int main()
{
	unsigned long A, B, tmp;
	int cnt = 1;
	cin >> A >> B;

	if (A > B) {
		tmp = A;
		A = B;
		B = tmp;
	}
	tmp = A;

	for (int i = 2; i*i <= tmp; i++) {
		
		if (A % i == 0) {
			if (B % i == 0) {
				cnt++;
			}
			while (A % i == 0) {
				A /= i;
			}
		}
	}
	if (A != 1 && B % A == 0)	cnt++;
	printf("%d", cnt);


	return 0;

}