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: Accepted)

#!/usr/bin/env python3

import sys
import math
# from string import ascii_lowercase, ascii_upper_case, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from copy import deepcopy                    # to copy multi-dimentional matrix without reference
# from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)


def main():
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79...e+308
    # inf = 2 ** 64 - 1               # (for fast JIT compile in PyPy) 1.84...e+19
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())
    
    
    n = ii()

    # naive, O(nlgn)
    # cnt = 0
    # for k in range(2, n + 1):
    #     tmp = n
    #     if tmp % k == 1:
    #         cnt += 1
    #     elif tmp % k == 0:
    #         while tmp >= k and tmp % k == 0:
    #             tmp //= k
    #         if tmp % k == 1:
    #             cnt += 1
    # print(cnt)

    class Eratos:
        def __init__(self, num):
            assert(num >= 1)
            self.table_max = num
            # self.table[i] は i が素数かどうかを示す (bool)
            self.table = [False if i == 0 or i == 1 else True for i in range(num+1)]
            for i in range(2, int(math.sqrt(num)) + 1):
                if self.table[i]:
                    for j in range(i ** 2, num + 1, i):    # i**2 からスタートすることで定数倍高速化できる
                        self.table[j] = False
            # self.table_max 以下の素数を列挙したリスト
            self.prime_numbers = [2] if self.table_max >= 2 else []
            for i in range(3, self.table_max + 1, 2):
                if self.table[i]:
                    self.prime_numbers.append(i)
        
        def is_prime(self, num):
            """
            >>> e = Eratos(100)
            >>> [i for i in range(1, 101) if e.is_prime(i)]
            [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
            """
            assert(num >= 1)
            if num > self.table_max:
                raise ValueError('Eratos.is_prime(): exceed table_max({}). got {}'.format(self.table_max, num))
            return self.table[num]
        
        def prime_factorize(self, num):
            """
            >>> e = Eratos(10000)
            >>> e.prime_factorize(6552)
            {2: 3, 3: 2, 7: 1, 13: 1}
            """
            assert(num >= 1)
            if int(math.sqrt(num)) > self.table_max:
                raise ValueError('Eratos.prime_factorize(): exceed prime table size. got {}'.format(num))
            factorized_dict = dict()    # 素因数分解の結果を記録する辞書
            candidate_prime_numbers = [i for i in range(2, int(math.sqrt(num)) + 1) if self.is_prime(i)]
            # n について、√n 以下の素数で割り続けると最後には 1 or 素数となる
            # 背理法を考えれば自明 (残された数が √n より上の素数の積であると仮定。これは自明に n を超えるため矛盾)
            for p in candidate_prime_numbers:
                # これ以上調査は無意味
                if num == 1:
                    break
                while num % p == 0:
                    num //= p
                    try:
                        factorized_dict[p]
                    except KeyError:
                        factorized_dict[p] = 0
                    finally:
                        factorized_dict[p] += 1
            if num != 1:
                factorized_dict[num] = 1
            return factorized_dict
    

    eratos = Eratos(int(math.sqrt(n)))
    d_1 = eratos.prime_factorize(n - 1)
    # n = k + 1, 2k + 1, 3k + 1, ...    
    cnt = 0
    tmp = 1
    for _, v in d_1.items():
        tmp *= (v + 1)
    tmp -= 1
    cnt += tmp
    d_2 = eratos.prime_factorize(n)
    # 全約数 x (除: 1) について (n // x) ≡ 1 (mod x) なら cnt += 1とすればいいのでは
    # 約数の個数は高々 2√n 個なので間に合いそう？
    L = [range(v + 1) for _, v in d_2.items()]
    power_list = list(product(*L))
    prime_list = list(d_2.keys())
    for elm in power_list:
        # elm は具体的な各素数の指数を指定したリストである
        div = 1
        for i in range(len(prime_list)):
            div *= pow(prime_list[i], elm[i])
        # print(f"div: {div}")
        if div != 1:
            assert(n % div == 0)
            copy_n = n
            while copy_n % div == 0:
                copy_n //= div
            if copy_n % div == 1:
                cnt += 1

    print(cnt)


if __name__ == "__main__":
    main()