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)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let v = read_vec::<i64>();
    let (a, b) = (v[0], v[1]);
    let mut n = gcd(a, b);

    let mut root_n = 1;
    while root_n * root_n <= n {
        root_n += 1;
    }
    let primes = get_primes(root_n as usize + 2);
    debug!(primes);

    let mut factors = vec![];
    for &p in primes.iter() {
        while n % p as i64 == 0 {
            factors.push(p as i64);
            n /= p as i64;
        }
    }

    if n != 1 {
        factors.push(n);
    }
    factors.push(1);
    debug!(factors);
    let s = factors.iter().collect::<HashSet<_>>();

    println!("{}", s.len());
}

fn get_primes(n: usize) -> Vec<usize> {
    let mut is_prime = vec![true; n + 1];
    let mut primes = Vec::new();
    is_prime[0] = false;
    is_prime[1] = false;

    for i in 2..n + 1 {
        if is_prime[i] {
            primes.push(i);
            let mut j = 2 * i;
            while j < n {
                is_prime[j] = false;
                j += i;
            }
        }
    }
    primes
}

fn factorize(mut num: usize, primes: &Vec<usize>) -> std::collections::HashMap<usize, usize> {
    // max_primes >= (num)^(1/2)
    let mut dict = std::collections::HashMap::new();
    for &p in primes.iter() {
        while num % p == 0 {
            *dict.entry(p).or_insert(0) += 1usize;
            num /= p;
        }
        if num == 1 {
            break;
        }
    }
    if num != 1 {
        dict.insert(num, 1);
    }
    dict
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn gcd(a: i64, b: i64) -> i64 {
    if b == 0 {
        return a;
    }
    gcd(b, a % b)
}