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 main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).ok();
    let mut it = buf.split_whitespace();
    let a = it.next().unwrap().parse::<u64>().unwrap();
    let b = it.next().unwrap().parse::<u64>().unwrap();
    let mut div : Vec<u64> = Vec::new();
    for i in 2..a {
        if i * i > a { break; }
        if a % i == 0 {
            if b % i == 0 { div.push(i) }
            if b % (a/i) == 0 { div.push(a/i) }
        }
    }
    if a > 1 && b % a == 0 { div.push(a); }
    div.sort();
    let mut res : Vec<u64> = Vec::new();
    for d in div {
        if res.iter().all(|x| d % x != 0) { res.push(d); }
    }
    println!("{}", res.len() + 1);
}