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)

#!/usr/bin/env python3
import math
n = int(input())
if n == 2:
    print(1)
    exit()
# n-1の約数を全列挙か、n の約数を全部探索して、n//k*p = 1+k*qになるかを確かめる
def divisor(n):
    res = [n]
    for i in range(2,1+math.ceil(math.sqrt(n))):
        if n % i == 0:
            res.append(i)
            if i != n//i:
                res.append(n//i)
    return res
ans = len(divisor(n-1))
# n の約数で全列挙
res = []
lis = divisor(n)
for i in lis:
    # n を i で割れるだけわる
    m = n
    while m % i == 0:
        m = m // i 
    if m % i == 1:
        ans += 1
        res.append(i)
        #print(m,i)
print(ans)