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)

import math

def divisor(num):
    if num < 0:
        return 0
    if num == 1:
        return [1, ]

    divisor_list = [1, ]
    divisor_list_r = [num, ]

    i = 2
    while i * i <= num:
        if num % i == 0:
            divisor_list.append(i)
            divisor_list_r.append(num//i)
        i += 1

    return divisor_list+divisor_list_r[::-1]


N = int(input())
ans = len(divisor(N-1))
cand = divisor(N)
for i, num in enumerate(cand):
    if i >= len(cand)//2:
        break
    if i == 0:
        ans += 1
        continue
    num2 = cand[-i]
    while num2 < num:
        tmp1, tmp2 = divmod(num2, num)
        if tmp2 == 0:
            num2 = tmp1
        else:
            num2 = tmp2
    if num2 == 1:
        ans += 1
print(ans)