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 PyPy3 (2.4.0) (judge verdict: Accepted)

N = int(input())

divisor = set([N, N - 1])
x = N
for i in range(2, int(x ** 0.5) + 1):
    z = x
    while z % i == 0:
        z //= i
    if z % i == 1:
        divisor.add(i)
        
    if i == x // i:
        continue
        
    z = x
    while z % (x // i) == 0:
        z //= (x // i)
    if z % (x // i) == 1:
        divisor.add((x // i))
        
x = N - 1
for i in range(2, int(x ** 0.5) + 1):
    if x % i == 0:
        divisor.add(i)
        divisor.add(x // i)

print(len(divisor) if N > 2 else 1)