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 PHP7 (7.0.15) (judge verdict: Time Limit Exceeded)

<?php
fscanf(STDIN, "%d %d", $a, $b);
$fixA = $a;
$fixB = $b;
$init = 2;
while ($a !== 1) {
$i = $init; while ($i < $fixA/2) {
if ($a % $i == 0) {
$resultA[] = $i;
$a /= $i;
break;
} $i++;
}
$init = $i; }
$resultA = array_unique($resultA);

$init = 2;
while ($b !== 1) {
$i = $init; while ($i < $fixB/2) {
if ($b % $i == 0) {
$resultB[] = $i;
$b /= $i;
break;
} $i++;
}
$init = $i; }
$resultB = array_unique($resultB);

$primeNumber = array_intersect($resultA, $resultB);

echo count($primeNumber) + 1;