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?

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.

12 18

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.

420 660

4

1 2019

1

$1$ and $2019$ have no positive common divisors other than $1$.