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::{self, BufRead};

fn main() {
    let stdin = io::stdin();
    let mut lines = stdin.lock().lines();
    let ab = lines.next().unwrap().unwrap();
    let mut ab = ab.split_whitespace();
    let a: i64 = ab.next().unwrap().parse().unwrap();
    let b: i64 = ab.next().unwrap().parse().unwrap();

    let mut n = gcd(a, b);
    let mut ans = 1;
    for i in 2.. {
        if i * i > n {
            if n > 1 {
                ans += 1;
            }
            break;
        }
        if n % i == 0 {
            while n % i == 0 {
                n /= i;
            }
            ans += 1;
        }
    }

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

fn gcd(a: i64, b: i64) -> i64 {
    let mut t = (a, b);
    while t.1 != 0 {
        t = (t.1, t.0 % t.1);
    }
    t.0
}