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: Runtime Error)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
use std::collections::HashMap;
use std::cmp::{max,min};
fn main(){
    input!{
        mut ab:[usize;2],
    }
    ab.sort();
    let (a,b)= (ab[0],ab[1]);
    let c = gcd(a, b);
    let ans = sieve_of_eratosthenes(c).into_iter().filter(|x| c % x == 0).count();
    println!("{}",ans+1);
}


fn is_prime(x:i64) -> bool {
    if x == 2 {return true;}
    if x < 2 || x % 2 == 0 {return false;}
    let mut j = 3;
    while j <= (x as f64).sqrt() as i64 {
        if x % j == 0 {
            return false;
        }
        j += 2;
    }
    true
}

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

fn sieve_of_eratosthenes(n: usize) -> Vec<usize> {
    let mut spf = vec![None; n+1];
    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);
            spf[i] = Some(i);
        }
        for prime in &primes {
            if i * prime >= n + 1 || prime > &spf[i].unwrap() {
                break;
            }
            is_prime[i * prime] = false;
            spf[i * prime] = Some(*prime);
        }
    }
    primes
}