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 std::io::Read;

fn is_prime(n: usize) -> bool {
    if n == 1 {
        return false;
    }

    for i in 2..n {
        if i * i > n {
            break;
        }
        if n % i == 0 {
            return false;
        }
    }
    return true;
}

fn main() {
    let mut buf = String::new();

    std::io::stdin().read_to_string(&mut buf).unwrap();

    let mut iter = buf.split_whitespace();
    let a: usize = iter.next().unwrap().parse().unwrap();
    let b: usize = iter.next().unwrap().parse().unwrap();

    let mut ans = 1;
    let mut i = 2;
    while i * i <= std::cmp::min(a, b) {
        if !is_prime(i) {
            i += 1;
            continue;
        } else if a % i == 0 && b % i == 0 {
            ans += 1;
        }

        i += 1;
    }

    // for i in 1..std::cmp::min(a, b) + 1 {
    //     if i * i > std::cmp::min(a, b) {
    //         break;
    //     }
    //
    //     if i == 1 {
    //         ans += 1;
    //     } else if !is_prime(i) {
    //         continue;
    //     } else if a % i == 0 && b % i == 0 {
    //         ans += 1;
    //     }
    // }

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