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: Time Limit Exceeded)

use std::io;

fn main() {
    let n: i64 = scan_input().trim().parse().unwrap();
    let size = (2..(n + 1)).into_iter()
        .map(|i| { calc(n, i) })
        .filter(move |num: &i64| { num == &1 })
        .count();
    println!("{}", size);
}

fn scan_input() -> String {
    let mut input = String::new();
    match io::stdin().read_line(&mut input) {
        Ok(_) => input,
        Err(_) => "".to_string(), 
    }
}

fn calc(n: i64, k: i64) -> i64 {
    if n < k { return n; }
    let param = match n % k {
        0 => n / k,
        _ => n - k
    };
    calc(param, k)
}