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: Time Limit Exceeded)

use std::f64;
use std::cmp::min;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn gcd(a: i64, b: i64) -> i64 {
    let mut numa :i64 = 0;
    let mut numb :i64 = 0;

    if a < b {
        numa = b;
        numb = a;
    } else if b < a {
        numa = a;
        numb = b;
    } else {
        return a;
    }
    if numb == 0 {
        return numa;
    }
    return gcd(numb, numa % numb);
}

fn is_prime(n: i64) -> bool {
    if (n == 0) || (n == 1) {
        return false;
    }

    let max = (n as f64).sqrt() as i32;

    for i in 2..max+1 {
        if n % i as i64 == 0 {
            return false;
        }
    }

    return true;
}

fn main() {
    input!{
        a: i64,
        b: i64,
    }

    let gcdnum = gcd(a, b);
    let mut n = 0;

    //let max = (min(a, b) as f64).sqrt() as i64;
    let max = gcdnum;
    let mut tmax = max;
    if is_prime(gcdnum) {
        println!("2");
    } else {
    for i in 2..max+1 {
        if gcdnum % i == 0 {
            if is_prime(i) {
                n = n + 1;
            }
            tmax = max / i;
        }
        if i > tmax {
            break;
        }
    }
    println!("{}", n+1);
    }
}