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)

import sys
sys.setrecursionlimit(10**7)
n = int(input())

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append((i, cnt))

    if temp!=1:
        arr.append((temp, 1))

    if arr==[]:
        arr.append((n, 1))

    return arr

def get_divisors(num):
    f_divs, l_divs = [], []
    for i in range(1, int(num**0.5)+1):
        if not num % i:
            f_divs.append(i)
            if i != num // i:
                l_divs.append(num // i)
    return f_divs + l_divs[::-1]

f = factorization(n)
num, cnt = zip(*f)
ans = len(get_divisors(n-1))-1

def dfs(idx, x):
    nx = x
    if idx >= len(num):
        if x == 1: return
        k = n
        while k % x == 0: k //= x
        if (k-1) % x == 0:
            global ans
            ans += 1
        return

    for i in range(cnt[idx]+1):
        dfs(idx+1, nx)
        nx *= num[idx]

dfs(0, 1)
print(ans)