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'
def nd(n)
  Prime.prime_division(n).map {|p, e| e + 1 }.inject(:*)
end
def _divisors(primes, k)
  Enumerator.new do |y|
    if primes.size == k
      y << 1
    else
      p, e = primes[k]
      _divisors(primes, k + 1).each {|d|
        (e + 1).times {|e1|
          y << p ** e1 * d
        }
      }
    end
  end
end
def d(n)
  _divisors(Prime.prime_division(n), 0).to_a.sort
end

n=gets.to_i
c=0
if n==2
  puts 1
else
  c=nd(n-1)
  d(n).each do |k|
    next if k==1||k==n
    m=n
    while m%k==0
      m/=k
    end
    c+=1 if m%k==1
  end
  puts c
end