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 input::*;
use std::cmp::*;
use std::collections::HashSet;
use std::io::*;
use std::num::*;
use std::str::*;

//#[rustfmt::skip]
mod input {
    use super::*;
    pub fn read<T: FromStr>() -> T {
        stdin()
            .bytes()
            .map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>()
            .parse::<T>()
            .ok()
            .unwrap()
    }
    pub fn string() -> String {
        read()
    }
    pub fn int() -> i64 {
        read()
    }
    pub fn char() -> char {
        read::<String>().pop().unwrap()
    }
    pub fn vecchar() -> Vec<char> {
        string().chars().collect()
    }
    pub fn vecint(n: i64) -> Vec<i64> {
        let mut vec = Vec::new();
        for i in 0..n {
            vec.push(int())
        }
        vec
    }
}
fn main() {
    let (a, b) = (int(), int());
    let (adiv, bdiv) = (divisor(a), divisor(b));
    let mut ans = 0;
    for ai in &adiv {
        for bi in &bdiv {
            if ai == bi {
                ans += 1
            }
        }
    }
    //println!("{:?}", adiv);
    //println!("{:?}", bdiv);
    println!("{}", ans + 1)
}

fn divisor(mut n: i64) -> HashSet<i64> {
    let mut r = HashSet::new();
    let mut l = 2;
    while l * l <= n {
        if n % l == 0 {
            n /= l;
            r.insert(l);
            continue;
        }
        l += 1;
    }
    r.insert(n);
    r
}