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)

use std::io;
use std::str::FromStr;

fn input_int_tuple() -> (u64, u64) {
  let mut line: String = String::new();
  io::stdin().read_line(&mut line);
  let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
  (vec[0], vec[1])
}

fn gcd(a: u64, b: u64) -> u64 {
  if b == 0 {
    a
  } else {
    gcd(b, a % b)
  }
}

fn divisors(n: u64) -> Vec<u64> {
  let mut divs: Vec<u64> = Vec::new();
  for i in 1..(n as f64).sqrt() as u64 + 1 {
    if n % i == 0 {
      divs.push(i);
      if i != (n / i) {
        divs.push(n/i);
      }
    }
  }
  divs.sort();
  divs
}

fn main() {
  let (a, b) = input_int_tuple();
  let gcd = gcd(a, b);
  let divs = divisors(gcd);
  let mut ret_vec: Vec<u64> = Vec::new();
  for n in divs {
    let mut flag = false;
    for m in &ret_vec {
      if *m == 1 {
        continue;
      }
      if n % *m == 0{
        flag = true;
        break;
      }
    }
    if flag == false {
      ret_vec.push(n);
    }
  }
  println!("{:?}", ret_vec.len());
}