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 OCaml (4.02.3) (judge verdict: Accepted)

let n = read_int ()

let rec factors m a acc =
  if a * a > m then acc
  else if a * a = m then a :: acc
  else if m mod a = 0 then factors m (a + 1) (a :: (m / a) :: acc)
  else factors m (a + 1) acc

let fs = factors (n - 1) 1 [] |> List.length

let rec test m k =
  if m = 0 then false else if m mod k = 0 then test (m / k) k else m mod k = 1

let fn =
  factors n 1 []
  |> List.filter (fun x -> x <> 1)
  |> List.map (test n)
  |> List.fold_left (fun acc x -> if x then acc + 1 else acc) 0

;;
print_int (fs - 1 + fn);
print_newline ()