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: Wrong Answer)

require "prime"
n = gets.to_i
a = (n-1).prime_division
b = n.prime_division
ans = a.map(&:last).inject(1){|acc,v| acc * (v+1)}

def solv(a,b)
  keys = a.map(&:first) & b.map(&:first)
  cnt = Hash.new(0)
  keys.each do |key|
    val = [a.assoc(key), b.assoc(key)].min
    cnt[key] = val
  end
  cnt
end

c = solv(a,b)
cans = c.values.inject(1){|acc,v| acc * (v+1)}
puts ans + cans