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'

n = gets.to_i
ans = (n-1).prime_division.inject(1){|r, (_, v)| r * (v + 1)} - 1

class Integer
    def my_divisor_list2
      return [1] if self == 1
      Prime.prime_division(self).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
    end
end

n.my_divisor_list2.each do |a|
    next if a == 1
    x = n
    x /= a while x % a == 0
    ans += 1 if x % a == 1
end

puts ans