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

@count2 = 0
@h2 = {}
def hoge2(t,arr, i)
  if i < 0
    if !@h[t] && t != 1
      @h[t] = true
      m = N
      m /= t while m % t == 0
      if m % t == 1
        @count += 1
      end
    end
  else
    hoge2(t,arr, i-1)
    hoge2(t*arr[i],arr, i-1)
  end
end
def try2(n)
  arr = Prime.prime_division(n).map { |a,i| [a] * i }.flatten
  hoge2(1,arr,arr.length-1)
  @count2
end


@count = 0
@h = {}
def hoge(t,arr, i)
  if i < 0
    if !@h[t] && t != 1
      @h[t] = true
      @count += 1
    end
  else
    hoge(t,arr, i-1)
    hoge(t*arr[i],arr, i-1)
  end
end
def try1(n)
  arr = Prime.prime_division(n).map { |a,i| [a] * i }.flatten

  hoge(1,arr,arr.length-1)
  @count
end

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