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::io::Read;

fn read<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

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

/// Enumerate divisors.
///
/// The time complexity is O(sqrt(n)).
///
/// Output: if n = 6, then {1, 2, 3, 6}.
fn divisors(n: u64) -> Vec<u64> {
    let mut res = Vec::new();
    let mut x = 1;
    while x * x <= n {
        if n % x == 0 {
            res.push(x);
            if x * x != n {
                res.push(n / x);
            }
        }
        x += 1;
    }
    res.sort();
    res
}

/// Perform prime factorization.
///
/// The time complexity is O(sqrt(n)).
///
/// Input: n is an integer greater than or equal to 2.
/// Output: if n = 540 = 2^2 * 3^3 * 5, then {(2, 2), (3, 3), (5, 1)}.
fn prime_factor(mut n: u64) -> Vec<(u64, u64)> {
    let mut res = Vec::new();
    let mut x = 2;
    while x * x <= n {
        let mut num = 0;
        while n % x == 0 {
            n /= x;
            num += 1;
        }
        if num != 0 {
            res.push((x, num));
        }
        x += 1;
    }
    if n != 1 {
        res.push((n, 1));
    }
    res
}

fn main() {
    let a: u64 = read();
    let b: u64 = read();
    let c = gcd(a, b);
    let primes = prime_factor(c);
    println!("{}", primes.len() + 1);
}