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: Runtime Error)

fn main() {
    let s = {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        s.trim_right().to_owned()
    };

    let (a, b) = {
        let mut ws = s.split_whitespace();
        let a: u64 = ws.next().unwrap().parse().unwrap();
        let b: u64 = ws.next().unwrap().parse().unwrap();
        (a, b)
    };

    // println!("{}", gcd(a, b));
    // println!("{:?}", sieve(a));

    let gcd = gcd(a, b);
    let primes = sieve(gcd);

    let mut count = 0;
    for x in primes {
        if gcd % x == 0 {
            count += 1;
        }
    }
    count += 1;
    println!("{}", count)
}

fn gcd(a: u64, b: u64) -> u64 {
    let c = a % b;
    if c == 0 {
        b
    } else {
        gcd(b, c)
    }
}

fn sieve(n: u64) -> Vec<u64> {
    let mut table: Vec<u64> = vec![0; (n + 1) as usize];
    let mut primes: Vec<u64> = Vec::new();

    for i in 2..n {
        if table[i as usize] == 0 {
            primes.push(i);
            for j in 2..n {
                if i * j > n {
                    break
                }
                table[(i * j) as usize] = 1
            }
        }
    }

    primes
}