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: Time Limit Exceeded)

use std::io::Read;
use std::collections::HashMap;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let a: usize = iterator.next().unwrap().parse().unwrap();
    let b: usize = iterator.next().unwrap().parse().unwrap();

    let gcd: usize = gcd(a, b);

    let pf = prime_factorization(gcd);

    let ans = pf.len() + 1;
    return ans.to_string();
}

fn gcd(mut x: usize, mut y: usize) -> usize {
    while y > 0 {
        let r = x % y;
        x = y;
        y = r;
    }
    return x;
}

fn prime_factorization(mut n: usize) -> HashMap<usize, usize> {
    let mut result: HashMap<_, _> = HashMap::new();

    let mut d = 2;
    while d <= n {
        while n % d == 0 {
            *result.entry(d).or_insert(0) += 1;
            n /= d;
        }
        d += 1;
    }

    return result;
}