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)]
use std::cmp;
use std::io::Write;

// Input macro from https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
    let mut iter = $s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
  ($($r:tt)*) => {
    let s = {
      use std::io::Read;
      let mut s = String::new();
      std::io::stdin().read_to_string(&mut s).unwrap();
      s
    };
    let mut iter = s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
}

macro_rules! input_inner {
  ($iter:expr) => {};
  ($iter:expr, ) => {};

  ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
    let $var = read_value!($iter, $t);
    input_inner!{$iter $($r)*}
  };
}

macro_rules! read_value {
  ($iter:expr, ( $($t:tt),* )) => {
    ( $(read_value!($iter, $t)),* )
  };

  ($iter:expr, [ $t:tt ; $len:expr ]) => {
    (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
  };

  ($iter:expr, chars) => {
    read_value!($iter, String).chars().collect::<Vec<char>>()
  };

  ($iter:expr, usize1) => {
    read_value!($iter, usize) - 1
  };

  ($iter:expr, [ $t:tt ]) => {{
    let len = read_value!($iter, usize);
    (0..len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
  }};

  ($iter:expr, $t:ty) => {
    $iter.next().unwrap().parse::<$t>().expect("Parse error")
  };
}

use std::collections::BTreeSet;
fn divisor(n: i64) -> BTreeSet<i64> {
  let mut i = 1;
  let mut div = BTreeSet::new();
  while i * i < n {
    if n % i == 0 {
      div.insert(i);
      if n / i != i {
        div.insert(n / i);
      }
    }
    i += 1;
  }
  div
}

fn main() {
  let out = std::io::stdout();
  let mut out = std::io::BufWriter::new(out.lock());
  macro_rules! puts {
    ($($format:tt)*) => (write!(out,$($format)*).unwrap());
  }
  input! {
    n: i64,
  }
  let dn = divisor(n);
  let mut ans = 0;
  for &d in &dn {
    if d == 1 {
      continue;
    }
    let mut x = n;
    while x % d == 0 {
      x /= d;
    }
    if x % d == 1 {
      ans += 1;
    }
  }
  let dm = divisor(n-1);
  for &d in &dm {
    if dn.contains(&d) {
      continue;
    }
    if n % d == 1 {
      ans += 1;
    }
  }

  puts!("{}\n", ans);
}