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: Accepted)

#![allow(unused_imports)]

use proconio::{input, fastout};
use proconio::marker::*;

#[fastout]
fn main() {
    input! {
    a: usize,
    b: usize
    }
    let da = divisor(a);
    let mut cd = vec![];
    for i in da {
        if b % i == 0 {
            cd.push(i);
        }
    }
    let mut ans = 0;
    for i in (0..cd.len()).rev() {
        let mut f = true;
        for j in 0..i {
            if gcd(&cd[i], &cd[j]) != 1 {
                f = false;
                break;
            }
        }
        if f { ans += 1; }
    }
    println!("{}", ans);
}

fn divisor(m: usize) -> Vec<usize> {
    let mut l = Vec::new();
    for i in 1.. {
        if i * i > m { break; }
        if m % i == 0 {
            l.push(i);
            l.push(m / i);
        }
    }
    l.sort();
    l.dedup();
    l
}

// O(log(N))
fn gcd(a: &usize, b: &usize) -> usize {
    let (mut x, mut y) = if a < b { (*b, *a) } else { (*a, *b) };
    let mut r = x % y;
    while r != 0 {
        x = y;
        y = r;
        r = x % y;
    }
    y
}