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>
#define ALL(a)  (a).begin(),(a).end()
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<long long, long long> Pll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long> vll;
typedef vector<vector<long long>> vvll;
template <typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template <typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
const int INT_INF = 1LL << 30;

bool is_OK(ll N, ll x){
  if(N == 1) return true;
  if(N < x) return false;

  if(N % x == 0) return is_OK(N/x, x);
  else if((N-1)%x != 0) return false;
  else return is_OK(N - N/x * x, x);
}

int main(){
  ll N;
  cin >> N;
  if(N == 2){
    cout << 1 << endl;
    return 0;
  }
  ll ans = 1; //Nは必ず答え

  //N-1の素因数分解
  ll tmp = 1;
  ll NN = N-1;
  for(ll i = 2; i*i <= NN; i++){
    if(NN%i) continue;
    if(NN%i == 0){
      ll counter = 0;
      while(NN%i == 0){
        NN /= i;
        counter++;
      }
      tmp *= counter + 1;
    }
  }
  if(NN != 1) tmp *= 2;
  tmp--;
  ans += tmp;

  //if(sqrt(N) * sqrt(N) == N) ans++; //平方数なら平方根は答え
  vll A(0); //平方根未満の約数
  for(ll i = 2; i*i < N; i++){
    if(N%i == 0) A.push_back(i);
  }
  for(int i = 0; i < A.size(); i++){
    if(is_OK(N, A.at(i))) ans++;
  }
  cout << ans << endl;
}