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;

fn is_prime(n: usize) -> bool {
    if n == 1 {
        return false;
    }

    for i in 2..n {
        if i * i > n {
            break;
        }
        if n % i == 0 {
            return false;
        }
    }
    return true;
}

fn main() {
    // 素数を全部ピックアップ。残りのうち素数で割り切れない物の数を加算。
    let mut buf = String::new();

    std::io::stdin().read_to_string(&mut buf).unwrap();

    let mut iter = buf.split_whitespace();
    let a: usize = iter.next().unwrap().parse().unwrap();
    let b: usize = iter.next().unwrap().parse().unwrap();

    let mut c: Vec<usize> = vec![];
    let mut d: Vec<usize> = vec![];
    let mut e: Vec<usize> = vec![];

    for i in 1..std::cmp::min(a, b) + 1 {
        if a % i == 0 && b % i == 0 {
            c.push(i);
        }
    }

    // println!("{:?}", c);

    for i in 0..c.len() {
        if is_prime(c[i]) {
            d.push(c[i]);
        } else {
            e.push(c[i]);
        }
    }

    // println!("{:?}", d);
    // println!("{:?}", e);

    let mut cnt = e.len() + d.len();

    for i in 0..e.len() {
        for j in d.iter() {
            if e[i] == 1 {
                continue;
            }
            if e[i] > *j {
                if e[i] % *j == 0 {
                    // print!("{} ", e[i]);
                    cnt -= 1;
                    break;
                }
            } else {
                if *j % e[i] == 0 {
                    // print!("{} ", e[i]);
                    cnt -= 1;
                    break;
                }
            }
        }
    }

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