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)

fn factor(mut n: usize) -> Vec<(usize, usize)> {
    let mut ret = vec![];
    for i in 2.. {
        if i * i > n {
            break;
        }
        let mut cnt = 0;
        while n % i == 0 {
            cnt += 1;
            n /= i;
        }
        if cnt != 0 {
            ret.push((i, cnt));
        }
    }
    if n != 1 {
        ret.push((n, 1));
    }
    ret
}

fn main() {
    proconio::input! {
        a: usize,
        b: usize,
    }
    let afactors = factor(a);
    let bfactors = factor(b);
    let mut h = std::collections::HashMap::<usize, usize>::new();
    for (num, _count) in afactors {
        *h.entry(num).or_insert(0) += 1;
    }
    for (num, _count) in bfactors {
        *h.entry(num).or_insert(0) += 1;
    }
    let ans = h.iter().filter(|(_key, count)| **count >= 2).count() + 1;
    println!("{}", ans);
}