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Contest: Task: Related: TaskC TaskE

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 $1$.