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 (7.3.0) (judge verdict: Accepted)

N = int(input())

def divisors(n):
    ret1 = []
    ret2 = []
    i = 1
    while i**2 <= N:
        if n%i == 0:
            ret1 += [i]
            ret2 += [n//i]
        i += 1
    return ret1 + ret2

ans1 = len(set(divisors(N-1))) - 1
ans2_cand = set(divisors(N))
ans2 = 0
for k in ans2_cand:
    if k == 1: continue
    n = N
    while n%k == 0:
        n = n//k
    if n%k == 1:
        ans2 += 1


print(ans1 + ans2)