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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use std::collections::HashSet;

fn main() {
    input! {
        a: usize,
        b: usize,
    }
    let set_a = prime_factorize(a); 
    let set_b = prime_factorize(b); 
    let intersection = set_a.intersection(&set_b);

    let mut count = 1;
    for _ in intersection {
        //println!("{}", i);
        count+=1;
    }
    println!("{}", count);
}

fn enum_divisiors(n: usize) -> HashSet<usize> {
    let mut res = HashSet::new();
    let mut i = 1;
    while i * i <= n {
        if n % i == 0 {
            res.insert(i);
            if n/i != i {
                res.insert(n/i);
            }
        }
        i+=1;
    }
    return res;
}

fn prime_factorize(mut n: usize) -> HashSet<usize> {
    let mut res = HashSet::new();
    let mut a = 2;
    while a * a <= n {
        if n % a != 0 {
            a+=1;
            continue;
        }
        let mut _ex = 0;
        while n % a == 0 {
            _ex+=1;
            n/=a;
        }
        res.insert(a);
        a+=1;
    }

    if n != 1 {
        res.insert(n);
    }
    return res;
}