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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int64_t N;
  cin >> N;

  unordered_set<int64_t> divs;
  // まず2以上であるN-1の約数を追加
  if(N-1 > 1){ divs.insert(N-1); }
  int64_t div= 2;
  while(div*div < N){
    if((N-1)%div==0){
      divs.insert(div);
      divs.insert((N-1)/div);
    }
    div++;
  }

  // Nの約数を(N自身とsqrt(N)以下のものを試して)追加
  divs.insert(N);
  div= 2;
  while(div*div <= N){
    int cnt= 0;
    int64_t tmp= N;
    while(tmp%div==0){
      cnt++;
      tmp/= div;
    }
    if(cnt > 0 && tmp%div==1){
      divs.insert(div);
    }

    div++;
  }
  
  cout << divs.size() << endl;
}