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: Wrong Answer)

use std::collections::HashSet;

fn gcd(a: u64, b: u64) -> u64 {
    let ab_max = std::cmp::max(a, b);
    let ab_min = std::cmp::min(a, b);
    let r = ab_max % ab_min;
    if r == 0 {
        ab_min
    } else {
        gcd(ab_min, r)
    }
}

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let a: u64 = iter.next().unwrap().parse().unwrap();
    let b: u64 = iter.next().unwrap().parse().unwrap();

    let mut ab_gcd = gcd(a, b);

    let mut divisor = HashSet::new();
    divisor.insert(1);

    let ab_sqrt = (ab_gcd as f64).sqrt() as u64 + 2;

    for i in 2..ab_sqrt {
        if ab_gcd == 1 {
            break;
        }

        while ab_gcd % i == 0 {
            divisor.insert(i);
            ab_gcd /= i;
        }
    }

    println!("{}", divisor.len());
}