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 (1.42.0) (judge verdict: Wrong Answer)

use proconio::{fastout, input};
use std::cmp;

#[fastout]
fn main() {
    input! {
        a: u64,
        b: u64,
    }
    let mut ans = 1;
    let mut is_prime = vec![true; 11e5 as usize];
    let mut i = 2usize;
    let c = cmp::max(a, b);
    while i * i <= c as usize {
        if !&is_prime[i] {
            i += 1;
            continue;
        }
        if a % i as u64 == 0 && b % i as u64 == 0 {
            ans += 1;
        }
        for j in (i..11e5 as usize).step_by(i) {
            is_prime[j] = false;
        }
        i += 1;
    }
    if a == b && a != 1 && ans == 1 {
        ans += 1;
    }
    println!("{}", ans);
}