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)

use std::collections::*;
use std::*;

macro_rules! test {
    ($($input:expr => $output:expr),* $(,)*) => {
        #[test]
        fn solve_test() {
            $(
                assert_eq!(solve($input), $output);
            )*
        }
    };
}
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

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

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

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

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

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

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

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

macro_rules! stdin {
    () => {{
        use std::io::Read;
        let mut s = String::new();
        std::io::stdin().read_to_string(&mut s).unwrap();
        s
    }};
}

fn main() {
    println!("{}", solve(&stdin!()));
}

fn solve(src: &str) -> String {
    input! {
        source = src,
        a: u64,
        b: u64
    }
    let mut a_factors = factors(a, a, b);
    let b_factors = factors(b, a, b);

    a_factors.extend(b_factors);

    (1 + a_factors.len()).to_string()
}

fn factors(mut n: u64, a: u64, b: u64) -> HashSet<u64> {
    let mut r = HashSet::<u64>::new();
    let mut d = 2;
    while (n >= d * d) {
        if n % d == 0 {
            n /= d;
            if a % d == 0 && b % d == 0 {
                r.insert(d);
            }
        }
        d += 1;
    }
    r
}

test! {
    "12 18" => "3",
    "420 660" => "4",
    "1 2019" => "1",
}

#[allow(dead_code)]
pub fn gcd(a: u64, b: u64) -> u64 {
    if b == 0 {
        a
    } else {
        gcd(b, a % b)
    }
}
#[allow(dead_code)]
pub fn lcm(a: u64, b: u64) -> u64 {
    a / gcd(a, b) * b
}
#[allow(dead_code)]
#[doc = " (gcd, x, y)"]
pub fn extgcd(a: i64, b: i64) -> (i64, i64, i64) {
    if b == 0 {
        (a, 1, 0)
    } else {
        let (gcd, x, y) = extgcd(b, a % b);
        (gcd, y, x - (a / b) * y)
    }
}
#[allow(dead_code)]
#[doc = " x ^ n % m"]
pub fn mod_pow(x: u64, n: u64, m: u64) -> u64 {
    let mut res = 1;
    let mut x = x % m;
    let mut n = n;
    while n > 0 {
        if n & 1 == 1 {
            res = (res * x) % m;
        }
        x = (x * x) % m;
        n >>= 1;
    }
    res
}
#[allow(dead_code)]
pub fn mod_inverse(a: u64, m: u64) -> u64 {
    let (_, x, _) = extgcd(a as i64, m as i64);
    ((m as i64 + x) as u64 % m) % m
}
#[allow(dead_code)]
pub fn fact_table(len: usize, m: u64) -> Vec<u64> {
    let mut res = vec![1; len + 1];
    for i in 1..len + 1 {
        res[i] = (i as u64 * res[i - 1]) % m;
    }
    res
}
#[allow(dead_code)]
#[doc = " Factorial and Inverse factorial table"]
pub fn fact_inv_table(size: usize, m: u64) -> (Vec<u64>, Vec<u64>) {
    let mut fact = vec![1; size];
    let mut fact_inv = vec![1; size];
    for i in 2..size {
        fact[i] = fact[i - 1] * i as u64 % m;
        fact_inv[i] = m - ((m / i as u64) * fact_inv[(m % i as u64) as usize] % m);
    }
    for i in 1..size {
        fact_inv[i] = fact_inv[i - 1] * fact_inv[i] % m;
    }
    (fact, fact_inv)
}
#[allow(dead_code)]
#[doc = " (a mod p, e when n! = a pe)"]
pub fn mod_fact(n: u64, p: u64, fact: &[u64]) -> (u64, u64) {
    if n == 0 {
        (1, 0)
    } else {
        let (a, b) = mod_fact(n / p, p, fact);
        let pow = b + n / p;
        if n / p % 2 != 0 {
            (a * (p - fact[(n % p) as usize]) % p, pow)
        } else {
            (a * fact[(n % p) as usize] % p, pow)
        }
    }
}
#[allow(dead_code)]
#[doc = " C(n, k) % p"]
pub fn mod_comb(n: u64, k: u64, p: u64, fact: &[u64]) -> u64 {
    if n < k {
        0
    } else {
        let (a1, e1) = mod_fact(n, p, fact);
        let (a2, e2) = mod_fact(k, p, fact);
        let (a3, e3) = mod_fact(n - k, p, fact);
        if e1 > e2 + e3 {
            0
        } else {
            a1 * mod_inverse(a2 * a3 % p, p) % p
        }
    }
}
#[allow(dead_code)]
#[doc = " H(n, k) % p"]
pub fn mod_comb_repetition(n: u64, k: u64, p: u64, fact: &[u64]) -> u64 {
    mod_comb(n - 1 + k, n - 1, p, fact)
}