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::cmp::{max, min};
use std::collections::HashSet;

fn gcd(a: usize, b: usize) -> usize {
    let mi = min(a, b);
    let ma = max(a, b);

    let r = ma % mi;
    if r == 0 {
        return mi;
    }

    gcd(mi, r)
}

fn solve_(e: usize, set: &mut HashSet<usize>) {
    for i in 2..((e as f64).sqrt() as usize + 2) {
        if e % i == 0 {
            set.insert(i);
            return solve_(e / i, set);
        }
        if i == ((e as f64).sqrt() as usize + 1) {
            set.insert(e);
        }
    }
}

fn solve(a: usize, b: usize) -> usize {
    let c = gcd(a, b);

    let mut set = HashSet::new();

    solve_(c, &mut set);

    set.len() + 1
}

#[test]
fn test() {
    assert_eq!(solve(420, 660), 4);
}

#[allow(unused_macros)]
macro_rules! parse_line {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            iter.next().unwrap().parse::<$t>().unwrap()
        }
    );

    ( $( $t:ty), +) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ( $(iter.next().unwrap().parse::<$t>().unwrap()),* )
        }
    );
}

#[allow(unused_macros)]
macro_rules! read_line {
    () => {{
        let mut line = String::new();
        ::std::io::stdin().read_line(&mut line).unwrap();
        line.pop();
        line
    }};
}

#[allow(unused_macros)]
macro_rules! parse_vec {
    ( $t:ty ) => {{
        let mut line = String::new();
        ::std::io::stdin().read_line(&mut line).unwrap();
        let iter = line.split_whitespace();
        iter.map(|v| v.parse::<$t>().unwrap()).collect::<Vec<_>>()
    }};
}

fn main() {
    let (a, b) = parse_line!(usize, usize);
    println!("{}", solve(a, b));
}