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::BTreeSet;
use std::io::prelude::*;
use std::io::stdin;
use std::str::FromStr;

fn main() {
    let mut a: u64 = read();
    let mut b: u64 = read();

    let mut a_divisor = BTreeSet::new();
    let mut b_divisor = BTreeSet::new();
    a_divisor.insert(1);
    b_divisor.insert(1);

    for i in 2..a.isqrt() + 1 {
        if a % i == 0 {
            a_divisor.insert(i);
        }
        while a % i == 0 {
            a /= i
        }
    }
    for i in 2..b.isqrt() + 1 {
        if b % i == 0 {
            b_divisor.insert(i);
        }
        while b % i == 0 {
            b /= i
        }
    }

    let intersection: Vec<_> = a_divisor.intersection(&mut b_divisor).collect();

    let mut ans = 0;
    'outer: for i in 0..intersection.len() {
        for j in i + 1..intersection.len() {
            if gcd(*intersection[j], *intersection[i]) != 1 {
                break 'outer;
            }
        }
        ans += 1;
    }

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

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

trait IntegerSqrt {
    fn isqrt(&self) -> Self;
}

impl IntegerSqrt for u64 {
    fn isqrt(&self) -> Self {
        if *self < 2 {
            *self
        } else {
            let s = (*self >> 2).isqrt() << 1;
            let l = s + 1;
            if l * l > *self {
                s
            } else {
                l
            }
        }
    }
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|b| b.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}