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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use std::collections::HashMap;

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

fn factorize(mut n:usize) -> HashMap<usize, bool> {
  let base_n = n;
  let mut map: HashMap<usize, bool> = HashMap::new();
  let mut index = 2;
  while index * index <= base_n {
    while n % index == 0 {
      map.insert(index, true);
      n /= index;
    }
    index += 1;
  }
  if n != 1 {
    map.insert(n, true);
  }
  map
}

fn main() {
  input! {
    a: usize,
    b: usize
  }
  
  println!("{}", factorize(gcd(a,b)).keys().len() + 1);
}