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

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let s = stdin
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok().unwrap()
}

fn string() -> String {
    read()
}

fn int() -> i64 {
    read()
}

fn char() -> char {
    read::<String>().pop().unwrap()
}

fn vecchar() -> Vec<char> {
    string().chars().collect()
}

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) = (div(a), div(b));

    let mut ans = 0;
    for ad in adiv {
        if bdiv.contains(&ad) {
            ans += 1;
        }
    }
    println!("{}", ans)
}

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