Problem:
Score : 600 points

Problem Statement

Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.

In how many choices of K will N become 1 in the end?

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of choices of K in which N becomes 1 in the end.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_imports)]pub mod pcl{}
fn divisors(n: i64) -> Vec<i64> {
    let mut res = Vec::new();
    for i in 1..((n as f64).sqrt() as i64 + 1) {
        if n % i == 0 {
            res.push(i);
            res.push(n / i);
        }
    }
    res.sort();
    res.retain(|&x| x != 1);
    res
}
fn main() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let n = s.trim().parse::<i64>().unwrap();
    let mut res = divisors(n - 1).len();
    let divs = divisors(n);
    for div in divs {
        let mut n = n;
        while n % div == 0 {
            n /= div;
        }
        if n % div == 1 {
            res += 1;
        }
    }
    println!("{}", res);
}