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: Wrong Answer)

import sys
sys.setrecursionlimit(10 ** 5)

from math import factorial as fact
n = int(input())
if n == 1:
  print(1)
  exit()

def f(n, k):
  if n < k:
    return n
  if n % k == 0:
    return f(n/k, k)
  else:
    return n % k
cnt = 0
p = {}
tmp = n - 1

for i in range(2, int(n**0.5)+1):
  if tmp % i == 0:
    while tmp % i == 0:
      tmp //= i
      p[i] = p.get(i, 0) + 1
if tmp != 1:
  p[tmp] = p.get(tmp, 0) + 1
ans = 1
for pp in p.values():
  ans *= pp + 1
ans -= 1
tmp = n

p = {}
for i in range(2, int(n**0.5)+1):
  if tmp % i == 0:
    while tmp % i == 0:
      tmp //= i
      p[i] = p.get(i, 0) + 1

for pp in p:
  if f(n, pp) == 1:
    ans += 1
      

print(ans + 1)