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)

fn read_stdin() -> String {
    use std::io::Read;
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input).unwrap();
    input
}

fn scan<'a, I: Iterator<Item = &'a str>, T: std::str::FromStr>(tokens: &mut I) -> T {
    tokens.next().unwrap().parse().ok().unwrap()
}

fn main() {
    let input = read_stdin();
    let tokens = &mut input.split_whitespace();
    let a: usize = scan(tokens);
    let b: usize = scan(tokens);
    let ps = prime_factors(a);
    let qs = prime_factors(b);
    let mut ans = 1;
    let mut i = 0;
    let mut j = 0;
    while i < ps.len() && j < qs.len() {
        if ps[i] == qs[j] {
            ans += 1;
            i += 1;
            j += 1;
        } else if ps[i] < qs[j] {
            i += 1;
        } else {
            j += 1;
        }
    }
    println!("{}", ans);
}

fn prime_factors(mut n: usize) -> Vec<usize> {
    let mut res = Vec::new();
    for d in 2.. {
        if n == 1 {
            break;
        } else if n < d * d {
            res.push(n);
            break;
        }
        if n % d == 0 {
            res.push(d);
            while {
                n /= d;
                n % d == 0
            } {}
        }
    }
    res
}