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: Wrong Answer)

// Practice AtCoder
// author:Leonardone @ NEETSDKASU

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 mut a = get!();
  let mut b = get!();
  let mut vs = vec![];
   
  for i in 1.. {
    if i*i > a || i*i > b {
      break;
    }
    if a % i != 0 || b % i != 0 {
      continue;
    }
    if i > 1 {
      vs.push(i);
    }
    
    while a > 1 && a % i == 0 {
       a /= i;
       if a > 1&&b % a == 0 {
          vs.push(a);
    } if i == 1 { break;}}
    
    while b > 1 && b % i == 0 {
       b /= i;
    if b > 1 && a % b == 0 { 
      vs.push(b);
    } if i == 1 {break;}}
  }
if a> 1 && b% a == 0 { vs.push(a);}
 if b > 1 && a% b==0 { vs.push(b);}
  vs.sort(); 
  vs.dedup();
  let mut flag = vec![false; vs.len()];
  let mut ans = 1;
  for i in 0..vs.len() {
     if flag[i] {
      continue;
     }
     ans += 1;
     for j in i+1..vs.len() {
       if vs[j] % vs[i] == 0 { 
        flag[j] = true;
       }
     }
  }
  println!("{}", ans);
}