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)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

pub fn prime_factor(n: u64) -> std::collections::HashMap<u64, u64> {
    if n < 2 {
        return std::collections::HashMap::new();
    }
    let mut res = std::collections::HashMap::new();
    let mut n = n;
    let mut i = 2;
    while i * i <= n {
        while n % i == 0 {
            let c = res.entry(i).or_insert(0);
            *c += 1;
            n /= i;
        }
        i += 1;
    }
    if n != 1 {
        res.insert(n, 1);
    }
    res
}

fn solve() {
    input! {
        a: u64, b: u64
    };

    let aset = prime_factor(a).keys().copied().collect::<HashSet<_>>();
    let bset = prime_factor(b).keys().copied().collect::<HashSet<_>>();
    let res = aset.intersection(&bset).count() + 1;
    println!("{}", res);
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}