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 f_division_or_substraction():
    N = int(input())

    def divisor_list(n):
        """n の正の約数のリスト"""
        ret = set()
        for k in range(1, int(n**0.5) + 1):
            if n % k == 0:
                ret.update({k, n // k})
        return sorted(ret)

    ans = (len(divisor_list(N - 1)) - 1)  # K >= 2 なので、-1 する
    for k in divisor_list(N):
        if k == 1:
            continue
        n = N
        while n % k == 0:
            n //= k
        ans += 1

print(f_division_or_substraction())