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: Wrong Answer)

# frozen_string_literal: true
require 'prime'

n = gets.to_i
m = [Math.sqrt(n).to_i, 2].max
mm = [Math.sqrt(n - 1).to_i, 2].max

def mod1?(n, k)
  x = n
  loop do
    x /= k
    break if x % k != 0
  end
  x % k == 1
end

mmf = mm ** 2 == n - 1
mf = m ** 2 == n
count = 1
(2..mm).each do |i|
  if ((n - 1) % i).zero?
    count += 2
    count -= 1 if i == mm && mmf
  end
end
(2..m).each do |i|
  if (n % i).zero?
    count += 1 if mod1?(n, i)
    if mod1?(n / i, i) && (i != m || !mf)
      count += 1
    end
  end
end

puts count