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)

n = gets.to_i

def divisor(n)
	array = []
	(1..Math.sqrt(n)).each do |i|
		if n % i == 0
			array << i
			array << n / i unless i ** 2 == n
		end
	end
	array
end

ans = 0
ans += divisor(n-1).size - 1
divisor(n).each do |i|
	next if i == 1
	temp = n
	while temp % i == 0
		temp /= i
	end
	ans += 1 if (temp - 1) % i == 0
end

puts ans