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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using Graph = vector<vector<int>>;
 
typedef long long ll;
typedef long double ld;
 
#define REP(i,a,n) for(int (i)=(a); (i)<(int)(n); (i)++)
#define rep(i,n) REP(i,0,n)
#define vec vector 

const ll INF = 2e9;
const ll large_P = 1e9 + 7;


int main(){
  ll N;
  cin >> N;
  
  ll res = 0;
    
  
  ll i = 2;
  ll j = 2;
  ll range = pow(N, 0.50) + 5;
  //cout << range << endl;
  while(range > i && N-1 > i){
    if((N)% i == 0 || (N-1)%i == 0){
      ll temp = N;
      while(temp%i == 0) temp /= i;
      temp %= i;
      
      if(temp == 1) {
        res++;
        //cout << i << endl;
      }
      
      temp = N;
      if((N)% i == 0) j = N/i;
      else j = (N-1)/i;
      while(temp%j == 0) temp /= j;
      temp %= j;
      
      if(temp == 1 && i*i != N && j > i && j < N-1) {
        res++;
        //cout << i << endl;
      }
    }
    i++;
  }
  
  
  res += 2; // K = N, N+1


  

  cout << res << endl;
}