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'
class Prime
  def each_divisor(n, &block)
    pf = prime_division(n)
    num = pf.inject(1) { |prod, (_, e)| prod * e.succ }
    Enumerator.new(num) { |y| __each_divisor__(1, pf, &y.method(:<<)) }.tap { |enum| enum.each(&block) }
  end
  
  private def __each_divisor__(d, ary, &block)
    return yield(d) if ary.empty?

    ary = ary.dup
    prime, exp = ary.pop
    (0..exp).each { __each_divisor__(d, ary, &block); d *= prime }
  end
end

n = gets.to_i

# substraction only or `n` itself
m = Prime.each_divisor(n-1).size

m += Prime.each_divisor(n).count do |z|
  next false if z <= 1 || z * z > n

  s, t = n, 0
  s, t = s.divmod(z) while t == 0
  t == 1
end

p m

load __FILE__ unless $stdin.eof?