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 Rust (1.15.1) (judge verdict: Accepted)

// Practice AtCoder
// author:Leonardone @ NEETSDKASU

// 解説PDF読後

fn main() {
  let mut stdin = String::new();
  std::io::Read::read_to_string(&mut std::io::stdin(), &mut stdin). unwrap ();
  let mut stdin = stdin.split_whitespace();
  let mut get = || stdin.next(). unwrap ();
  macro_rules! get {
    ($t:ty) => (get().parse::<$t>(). unwrap ());
    () => (get!(i64));
  }
  let a = get!();
  let b = get!();

  let fa = factors(a);
  let fb = factors(b);
  let mut ans = 0;
  for f in fb {
      if fa.binary_search(&f).is_ok() { ans+=1;}
  }
  println!("{}", ans);
}

fn factors(v: i64) -> Vec<i64> {
  let mut ret = vec![1];
  let mut v = v;
  for i in 2.. {
     if i*i > v { break; }
     if v % i == 0 {
       ret.push(i);
       while  v % i == 0 { v /= i; }
     }
  }
  if v > 1 { ret.push(v); }
  ret
}