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::{stdin, Read, StdinLock};
use std::str::FromStr;
use std::iter::FromIterator;
use std::collections::HashSet;

fn main() {
    let cin = stdin();
    let mut scan = Scanner::new(cin.lock());
    let a = scan.read::<i64>();
    let b = scan.read::<i64>();
    let ad = divisors(a).into_iter().collect::<HashSet<_>>();
    let bd = divisors(b).into_iter().collect::<HashSet<_>>();
    let cd = &ad & &bd;
    let mut cdv = cd.into_iter().collect::<Vec<_>>();
    cdv.sort();
    let mut ret = vec![];
    'outer: for &c in &cdv {
        for &r in &ret {
            if r != 1 && c % r == 0 {
                continue 'outer;
            }
        }
        ret.push(c);
    }
    println!("{}", ret.len());
}

#[allow(dead_code)]
fn divisors(n: i64) -> Vec<i64> {
    let mut ret = vec![];
    for i in (1..).take_while(|i| i * i <= n) {
        if n % i == 0 {
            ret.push(i);
            if i != n / i {
                ret.push(n / i);
            }
        }
    }
    ret
}


// region template
#[allow(dead_code)]
fn iu(i: i64) -> usize { i as usize }

#[allow(dead_code)]
fn ui(i: usize) -> i64 { i as i64 }

#[allow(dead_code, deprecated)]
fn join<T: std::fmt::Display>(slice: &[T], sep: &str) -> String {
    let strings = slice.iter().map(|t| format!("{}", t)).collect::<Vec<_>>();
    strings.connect(sep)
}

#[allow(dead_code)]
fn arr<'a, S, T>(n: usize, mut f: S) -> Vec<T> where S: FnMut(usize) -> T + 'a {
    (0..n).map(|i| f(i)).collect::<Vec<T>>()
}

struct Scanner<'a> {
    cin: StdinLock<'a>,
}

#[allow(dead_code)]
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }

    fn read1<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }

    fn read<T: FromStr>(&mut self) -> T {
        self.read1().unwrap()
    }

    fn readn<T: FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read::<T>()).collect()
    }

    fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}
// endregion