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: Time Limit Exceeded)

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}

fn gcd(x: u64, y: u64) -> u64 {
    if x == 0 {
        y
    } else {
        gcd(y % x, x)
    }
}

pub fn primes(x: u64) -> Vec<u64> {
    match x {
        n if n < 2 => vec![],
        n if n == 2 => vec![2],
        n => {
            let mut ps = Vec::new();
            let mut target: Vec<u64> = (2..(n + 1)).collect();
            let mut i = 2;
            while i * i <= n {
                ps.push(i);
                target = target
                    .iter()
                    .filter(|&x| x % i != 0)
                    .cloned()
                    .collect::<Vec<u64>>();
                i = target[0];
            }
            ps.append(&mut target);
            ps
        }
    }
}

fn is_prime(x: u64)->bool {
    match x {
        x if x == 2 => true,
        x if x < 2 || x % 2 == 0 => false,
        _ => {
            let mut i = 3;
            while i * i <= x {
                if x % i == 0 {
                    return false;
                }
                i = i + 2;
            }
            true
        }
    }
}

use std::collections::HashSet;
fn div(n: u64) -> u64 {
    let ps = primes(1000000);
    let mut m = n;
    let mut hs = HashSet::new();
    for p in ps {
        while m % p == 0 {
            hs.insert(p);
            m /= p;
        }

        if p * p > n && n != 1 {
            hs.insert(n);
            break;
        }
        if is_prime(m) {
            hs.insert(m);
            break;
        }
        if m == 1 {
            break;
        }
    }
    //println!("{}, {:?}", n, hs);
    hs.len() as u64
}

fn pfn(n: u64) -> u64 {
    let m = n as usize;
    let mut v = vec![1u64; m + 1];
    let mut i:usize = 2;
    let mut j:usize = 2;
    while i * i < m {
        if v[i] == 1 {
            while i * j < m {
                v[i*j] = v[i] + v[j];
                j += 1;
            }
        }
        i += 1;
    }
    println!("{:?}", v);
    let ret = v[m];
    ret
}

fn main() {
    let v = read_vec::<u64>();
    let a = v[0];
    let b = v[1];

    let g = gcd(a, b);
    let ans = div(g);
    println!("{}", ans+1);
}