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)

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])
    if arr==[]:
        arr.append([n, 1])
    return arr
  
n = int(input())
a = factorization(n)
ans = 1
b = factorization(n-1)
chk = 1
for i in range(len(b)):
  chk *= (b[i][1]+1)
ans += chk-1
for i in range(len(a)):
    for j in range(1,a[i][1]+1):
      cnt = 0
      cnt += a[i][1]-a[i][1]%j
      #print(a[i][0]**j,cnt,n/(a[i][0]**cnt),n/(a[i][0]**cnt) % (a[i][0]**(j)))
      if n/(a[i][0]**cnt) % (a[i][0]**(j)) == 1:
            ans += 1
if len(a) == 1:
  ans -= 1
if n == 2:
  ans = 1