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)

from math import sqrt


def divisors(x):
    for i in range(1, int(sqrt(x)) + 1):
        if x % i == 0:
            yield i
            if x // i != i:
                yield x // i


n = int(input())

ans = 0
for e in divisors(n):
    if e == 1:
        continue

    r = n
    while r % e == 0:
        r //= e

    if r % e == 1:
        ans += 1

ans += len(list(divisors(n - 1))) - 1

print(ans)