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 (2.7.1) (judge verdict: Runtime Error)

require "prime"

# n - 1 の約数で、１以外であれば題意を満たす
def solve_1(n)
  Prime.prime_division(n-1).map(&:last).map(&:succ).inject(:*)-1
end

# nの約数について、割りきれなくなるまで破った後、mod==1なら題意を満たす
# また、n,n-1の両方に含まれる約数は無い
def solve_2(n)
  a = []
  1.upto(n) do |i|
    break if i*i > n
    if n % i == 0
      a << i
      a << n/i if i != n/i
    end
  end
  a = a - [1]
  ans = 0
  a.each do |i|
    m = n
    m /= i while m % i == 0
    if m % i == 1
      ans += 1
    end
  end
  ans
end

n = gets.to_i
ans = solve_1(n) + solve_2(n)
puts ans