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 PHP (5.6.30) (judge verdict: Accepted)

<?php
fscanf(STDIN,"%d %d",$a,$b);

$a_arr=yakusu($a);
$b_arr=yakusu($b);

$arr=array_intersect($a_arr,$b_arr);
$val=array_unique($arr);
asort($val);
$n_val=array_values($val);
//print_r($val);
$cnt=0;
for($i=0;$i<count($n_val);$i++){
	if(is_prime($n_val[$i])){
		$cnt++;
	}
}
echo $cnt+1;

function yakusu($target) {
	$limit = ceil(sqrt($target));
	for($i=1;$i<=$limit;$i++) {
		if($target%$i == 0) {
			$divisor_array[] = $i;
			$j = $target/$i;
			if ($i!=$j) {
				$divisor_array[] = $j;
			}
		}
	}
	return $divisor_array;
}

function is_prime($n){
    if($n < 2) {
        return false;
    } else if ($n == 2) {
        return true;
    }
    if($n % 2 == 0) {
        return false;
    }
    for ($i = 3; $i * $i <= $n; $i += 2) {
        if ($n % $i == 0) {
            return false;
        }
    }
    return true;
}
?>