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 Ruby (2.3.3) (judge verdict: Accepted)

require 'prime'

# 約数リスト
def divisors(n)
  n.prime_division.inject([1]) do |array, (k, v)|
    (0..v).map{ |i| k ** i }.product(array).map{ |a, b| a * b }
  end.sort
end

n = gets.to_i
divs = divisors(n)
prime_divs = Prime.prime_division(n)
ans = divisors(n - 1)
divs.each do |d|
  dup_n = n
  while true
    div, mod = dup_n.divmod(d)
    if mod == 1
      ans << d
      break
    elsif mod == 0
      break if dup_n == div
      dup_n = div
    else
      break
    end
  end
end
ans.delete(1)
puts ans.uniq.size