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: Wrong Answer)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;

fn calc_gcd(a: i64, b: i64) -> i64 {
    if b == 0 {
        return a;
    } else {
        return calc_gcd(b, a % b);
    }
}

fn main() {
    input! {
        A: i64,
        B: i64,
    }

    let gcd = calc_gcd(A, B);

    if gcd == 1 {
        println!("1");
        return;
    }

    let mut ans = 1;
    let mut N = gcd;
    for i in 2..((gcd as f64).sqrt() as i64 + 2) {
        if N % i == 0 {
            while N % i == 0 {
                N /= i;
            }
            ans += 1;
        }
    }

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