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: Accepted)

use std::io::Read;

pub fn read_stdin() -> String {
    let mut s = String::new();
    ::std::io::stdin().read_to_string(&mut s).unwrap();
    s
}

pub trait Parse {
    fn to<T>(&mut self) -> T
    where
        T: std::str::FromStr;
}

impl<'a> Parse for std::str::SplitWhitespace<'a> {
    fn to<T>(&mut self) -> T
    where
        T: std::str::FromStr,
    {
        let word = self.next().unwrap();
        word.parse().ok().unwrap()
    }
}

fn factorization(x: u64) -> Vec<u64> {
    let mut base = x;
    let mut div = 2;
    let mut v = Vec::new();
    if x != 1 {
        v.push(1)
    };

    let last = (base as f64).sqrt() as u64;
    while last >= div {
        if base % div == 0 {
            base /= div;
            v.push(div);
            continue;
        }
        div += 1;
    }
    v.push(base);
    v
}


fn main() {
    let source = read_stdin();
    let mut iter = source.split_whitespace();
    let a = iter.to();
    let b = iter.to();
    let mut aa = factorization(a);
    aa.dedup();
    let mut bb = factorization(b);
    bb.dedup();
    aa.extend(&bb);
    aa.sort();

    if aa.len() <= 1 {
        println!("{}", aa.len());
        return;
    }

    let mut ans = Vec::new();
    for w in aa.windows(2) {
        if w[0] == w[1] {
            ans.push(w[0]);
        }
    }
    ans.dedup();

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