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: Accepted)

use std::io::{self, BufRead};

pub fn find_divisors(n: i64) -> Vec<i64> {
  let mut res = Vec::new();
  let mut i = 1;
  while i * i <= n {
    if n % i == 0 {
      res.push(i);
      if i != n / i {
        res.push(n / i);
      }
    }
    i += 1
  }
  res
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let n: i64 = lines.next().unwrap().unwrap().parse().unwrap();

  let num_divs_n = find_divisors(n-1).len() - 1;
  let divs = find_divisors(n);
  let mut num_divs_n_1 = 0;
  for k in divs {
    if n % k != 0 {
      continue
    }
    let mut m = n;
    while k != 1 && m != 1 && m % k == 0 {
      m /= k;
    }
    if m % k == 1 {
      num_divs_n_1 += 1;
    }
  }

  println!("{}", num_divs_n_1 + num_divs_n);
}