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)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let (A, B): (usize, usize) = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        let mut iter = line.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap()
        )
    };

    const N: usize = 1_000_000;
    let mut p = vec![true; N + 1];
    for i in 2..N {
        if p[i] {
            let mut j = 2 * i;
            while j <= N {
                p[j] = false;
                j += i;
            }
        }
    }
    fn gcd(a: i64, b: i64) -> i64 { if b == 0 { a } else { gcd(b, a % b) } }
    let mut g = gcd(A as i64, B as i64) as usize;
    let ds = (1..(N + 1)).filter(|&i| A % i == 0 && B % i == 0 && p[i]).collect::<std::collections::HashSet<usize>>();
    for &d in &ds { while d > 1 && g % d == 0 { g /= d;} }
    let ans = if g > N { ds.len() + 1 } else { ds.len() };

    println!("{}", ans);
}