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: Accepted)

def make_divisors(n):
    divisors = [n]
    for i in range(2, int(n ** 0.5) + 1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n // i)

    divisors.sort()
    return divisors


def main():
    N = int(input())
    if N == 2:
        print(1)
        exit()
    n = make_divisors(N)
    m = make_divisors(N - 1)
    answer = m[:]
    cnt = len(answer)
    for i in n:
        tmp = N
        while tmp % i == 0:
            tmp //= i
        if tmp % i == 1:
            cnt += 1
            answer.append(i)
    answer.sort()
    print(cnt)


if __name__ == '__main__':
    main()