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: Time Limit Exceeded)

require 'prime'

N=gets.to_i

def try2(n)
  divs = Prime.prime_division(n)
  arr = divs.map { |a,i| [a] * i }.flatten

  count = 0
  i = 1
  h = {}
  while i < 2**arr.length
    t = 1
    arr.length.times do |j|
      if ((i >> j) & 1) == 1
        t *= arr[j]
      end
    end
    if !h[t]
      h[t] = true
      m = n
      m /= t while m % t == 0
      if m % t == 1
        count += 1
      end
    end
    i += 1
  end
  count
end

def try1(n)
  divs = Prime.prime_division(n)
  arr = divs.map { |a,i| [a] * i }.flatten

  count = 0
  i = 1
  h = {}
  while i < 2**arr.length
    t = 1
    arr.length.times do |j|
      if ((i >> j) & 1) == 1
        t *= arr[j]
      end
    end
    if !h[t]
      h[t] = true
      count += 1
    end
    i += 1
  end
  count
end

p (try2(N) + try1(N-1))