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)

n = gets.to_i

require 'prime'
def divisor_list x
  return [1] if x == 1
  Prime.prime_division(x).map do |e|
    Array.new(e[1]+1).map.with_index do |element, i|
      e[0]**i
    end
  end.inject{|p,q| p.product(q)}.map do |a|
    [a].flatten.inject(&:*)
  end.sort
end

divisors = divisor_list(n-1) + divisor_list(n)
results = [2, n , *divisors].uniq

def is_result? n, x
  return false if x < 2
  nn = n
  while(nn % x == 0) do
    nn = nn / x
  end
  nn = nn % x
  if nn == 1
    true
  else
    false
  end
end
puts results.find_all{|z| is_result?(n, z) }.size