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)

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 gcd(x: usize, y: usize) -> usize {
    // return y ? gcd(y, x % y) : x;
    if y == 0 {
        return x;
    }
    return gcd(y, x % y);
}

fn factorize(n: &mut usize) -> Vec<(usize, usize)> {
    let mut res: Vec<(usize, usize)> = vec![];
    for i in 2..*n {
        if (i * i) >= *n {
            break;
        }

        if *n % i != 0 {
            continue;
        }

        // res.push((i, 0));
        let mut cnt = 0;
        while *n % i == 0 {
            (*n) /= i;
            cnt += 1;
            // res.last().unwrap().1 += 1;
        }
        res.push((i, cnt));
    }
    if *n != 1 {
        res.push((*n, 1));
    }
    return res;
}

fn main() {
    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 g: usize = gcd(a, b);
    let f = factorize(&mut g);
    // println!("{:?}", f);
    println!("{}", f.len() + 1);

    // for i in 2..std::num::sqrt(g) {}

    // let mut ans = 1;
    // let mut i = 2;
    //
    // while (i * i) <= std::cmp::max(a, b) {
    //     if !is_prime(i) {
    //         i += 1;
    //         continue;
    //     } else if a % i == 0 && b % i == 0 {
    //         ans += 1;
    //     }
    //
    //     i += 1;
    // }
    //
    // println!("{}", ans);
}