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 Python (3.4.3) (judge verdict: Accepted)

import math

N = int(input())

def enumerate_divisors(N):
  all_divisors = set()
  for divisor in range(1, int(math.sqrt(N)) + 1):
    if N%divisor == 0:
      if divisor == 1:
        all_divisors.add(N/divisor)
      else:
        all_divisors = all_divisors | {divisor, N/divisor}

  all_divisors = list(all_divisors)
  return all_divisors

def calculate_reminder(N, d):
  while True:
    reminder = N%d
    if reminder == 0:
      N /= d
    else:
      return reminder

if N == 2:
  counter = 1
else:
  counter = len(enumerate_divisors(N-1))
  for div in enumerate_divisors(N):
    if calculate_reminder(N, div) == 1:
      counter += 1

print(counter)