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
  
def divisorize(n):
    ## 約数
    f = factorization(n)
    divisors = [1]
    for k in range(len(f)):
        num_divisors = len(divisors)
        for i in range(1, f[k][1] + 1):
            for j in range(num_divisors):
                divisors.append(divisors[j] * f[k][0] ** i)
    return sorted(divisors)
  
n = int(input())
ans = 1
b = factorization(n-1)
chk = 1
for i in range(len(b)):
  chk *= (b[i][1]+1)
ans += chk-1
#print(divisorize(n))
for i in divisorize(n)[1:]:
  m = n
  while True:
    if m < i:
      break
    if m%i == 0:
      m = m//i
    else:
      m = m%i
      if m == 1:
        ans += 1
        break
        
if n == 2:
  ans = 1
  
print(ans)