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: Runtime Error)

// Original: https://github.com/tanakh/competitive-rs
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

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

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

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

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

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

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

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

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

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

mod util {
    #[allow(dead_code)]
    pub fn chmin<T>(x: &mut T, y: T) -> bool
    where
        T: PartialOrd + Copy,
    {
        *x > y && {
            *x = y;
            true
        }
    }

    #[allow(dead_code)]
    pub fn chmax<T>(x: &mut T, y: T) -> bool
    where
        T: PartialOrd + Copy,
    {
        *x < y && {
            *x = y;
            true
        }
    }

    /// 整数除算切り上げ
    #[allow(dead_code)]
    pub fn roundup(a: i64, b: i64) -> i64 {
        (a + b - 1) / b
    }

    #[allow(dead_code)]
    pub fn ctoi(c: char) -> i64 {
        c as i64 - 48
    }
}

#[allow(unused_imports)]
use util::*;

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, VecDeque};

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

    // 素数かつ最大公約数を割り切れる数値の個数が答え
    let g = num::gcd(a, b);
    let mut ans = 1;
    let n = g as usize + 1;
    let mut sieve = vec![true; n];
    for i in 2..n {
        if !sieve[i] {
            continue;
        }
        for j in 2..n {
            if i * j >= n {
                break;
            }
            sieve[i * j] = false;
        }

        if g % i as i64 != 0 {
            continue;
        }

        if sieve[i] {
            ans += 1;
        }
    }

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

pub mod num {
    use std::ops::{Add, Div, Mul, Neg, Rem, Sub};
    pub trait NumOps<Rhs = Self, Output = Self>:
        Add<Rhs, Output = Output>
        + Sub<Rhs, Output = Output>
        + Mul<Rhs, Output = Output>
        + Div<Rhs, Output = Output>
        + Rem<Rhs, Output = Output>
        + Neg<Output = Output>
    {
    }

    impl<T, Rhs, Output> NumOps<Rhs, Output> for T where
        T: Add<Rhs, Output = Output>
            + Sub<Rhs, Output = Output>
            + Mul<Rhs, Output = Output>
            + Div<Rhs, Output = Output>
            + Rem<Rhs, Output = Output>
            + Neg<Output = Output>
    {
    }

    pub trait Num: PartialEq + NumOps + From<i32> {}

    impl Num for i32 {}
    impl Num for i64 {}

    /// Greatest common divisor
    pub fn gcd<T: Num + Copy>(a: T, b: T) -> T {
        if b == 0.into() {
            a
        } else {
            gcd(b, a % b)
        }
    }

    /// Least common multiple
    pub fn lcm<T: Num + Copy>(a: T, b: T) -> T {
        let g = gcd(a, b);
        a / g * b
    }
}