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: Wrong Answer)

import math
n = int(input())

# 1以外のNの約数
divisors = []
for i in range(1, int(math.sqrt(n)) + 1):
    if n % i == 0:
        divisors.append(i)
        divisors.append(n // i)
divisors.sort()

candidates = set()
for k in divisors[1:]:
    tmp = n
    while True:
        if tmp % k != 0:
            break
        tmp = tmp // k
    if (tmp - 1) % k == 0:
        candidates.add(k)

# N-1の約数
divisors = []
for i in range(1, int(math.sqrt(n-1)) + 1):
    if (n-1) % i == 0:
        divisors.append(i)
        divisors.append((n-1) // i)

for k in divisors[1:]:
    candidates.add(k)

print(len(candidates))