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)

fn read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn f(a: i64) -> Vec<i64> {
    let mut av = Vec::new();
    for i in 2..(a as f64).sqrt().ceil() as usize + 1 {
        if a % i as i64 == 0 {
            av.push(i as i64);
        }
    }
    let mut av2 = Vec::new();
    while !av.is_empty() {
        let n = av[0];
        av2.push(n);
        let mut i = 0;
        while i != av.len() {
            if av[i] % n == 0 {
                av.remove(i);
            } else {
                i += 1;
            }
        }
    }
    av2
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let a: i64 = read(&mut stdin_lock, &mut buf, b' ');
    let b: i64 = read(&mut stdin_lock, &mut buf, b'\n');

    let av = f(a);
    let bv = f(b);
    let mut ans = 1;
    for i in 0..av.len() {
        for j in 0..bv.len() {
            if av[i] == bv[j] {
                ans += 1;
            }
        }
    }
    println!("{}", ans);
}