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::collections::HashSet;
use std::io::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read_option<T: FromStr>() -> Option<T> {
    let stdin = 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()
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let opt = read_option();
    opt.expect("failed to parse token")
}

fn gcd(a: i64, b: i64) -> i64 {
    let mut a = a;
    let mut b = b;
    if a < b {
        std::mem::swap(&mut a, &mut b);
    }
    if b == 0 {
        return a;
    }
    let mut r;
    while (a % b) != 0 {
        r = a % b;
        a = b;
        b = r;
    }
    b
}
#[test]
fn test_gcd() {
    assert_eq!(gcd(10, 5), 5);
    assert_eq!(gcd(20, 30), 10);
    assert_eq!(gcd(100, 3), 1);
    assert_eq!(gcd(3, 60), 3);
    assert_eq!(gcd(30, 20), 10);
}

fn main() {
    let a: i64 = read();
    let b: i64 = read();
    let gcd = gcd(a, b);
    let mut set: HashSet<i64> = HashSet::new();
    let sqrt = (gcd as f64).sqrt().ceil() as usize;
    let mut now = gcd;
    for i in 2..sqrt + 1 {
        let i = i as i64;
        while now % i == 0 {
            set.insert(i);
            now = now / i;
        }
        if now == 1 {
            break;
        }
    }
    // dbg!(gcd, now);
    if gcd != 1 && now == gcd {
        set.insert(gcd);
    }
    // dbg!(gcd, sqrt, set.clone());
    println!("{}", set.len() + 1);
}