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)

#ifdef LOCAL
    #define _GLIBCXX_DEBUG
    #define __clock__
#else
    #pragma GCC optimize("Ofast")
#endif
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<ll>;
using VV = vector<VI>;
using VS = vector<string>;
using PII = pair<ll, ll>;

// tourist set
template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
  return '"' + s + '"';
}

string to_string(const char* s) {
  return to_string((string) s);
}

string to_string(bool b) {
  return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}

template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}

template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << '\n'; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
// tourist set end

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,b) FOR(i, 0, b)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<'\n'
#define p2(s, t) cout << (s) << " " << (t) << '\n'
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << '\n'
#define SZ(x) ((int)(x).size())
#define SORT(A) sort(ALL(A))
#define RSORT(A) sort(ALL(A), greater<ll>())
#define MP make_pair
#define p_yes() p("Yes")
#define p_no() p("No")

ll SUM(VI& V){
  return accumulate(ALL(V), 0LL);
}

ll MIN(VI& V){return *min_element(ALL(V));}
ll MAX(VI& V){return *max_element(ALL(V));}

void print_vector(VI& V){
  ll n = V.size();
  rep(i, n){
    if(i) cout << ' ';
    cout << V[i];
  }
  cout << endl;
}

ll gcd(ll a,ll b){
    if(b == 0) return a;
    return gcd(b,a%b);
}

ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b;
}

// long double
using ld = long double;
#define EPS (1e-14)
#define equals(a,b) (fabs((a)-(b)) < EPS)

void no(){p_no(); exit(0);}
void yes(){p_yes(); exit(0);}

const ll mod = 1e9 + 7;
const ll inf = 1e18;
const double PI = acos(-1);

ll N; 
    
// 素因数分解
// その素因数が何個あるかのmapを返す
map<ll, ll> factorize(ll n){
    map<ll, ll> mp;
    ll sq = sqrt(n);
    FOR(i, 2, sq+1){
        while(n%i==0){
            mp[i]++;
            n/=i;
        }
    }
    // 残り
    if(n!=1){
        mp[n]++;
    }
    return mp;
}

// 約数の種類数
// 6 => 1, 2, 3, 6なので4
ll yaku(ll n){
  auto mp = factorize(n);
  ll ret = 1;
  for(auto pa : mp){
    ret *= (pa.second+1);
  }
  return ret;
}

// 素数判定
bool isPrime(ll n){
    if (n < 2) return false;
    else if (n == 2) return true;
    else if (n % 2 == 0) return false;

    double sqrtNum = sqrt(n);
    for (ll i=3; i<=sqrtNum; i+=2){
        if (n%i == 0){
            return false;
        }
    }
    return true;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    cin>>N;

    auto mp = factorize(N-1);
    debug(mp);

    ll y = yaku(N-1);
    debug(y);
    ll ans;
    
    if(isPrime(N)){
      ans = y;
    }else{
      ans = y+1;
    }
    p(ans);
    return 0;
    
    ll cnt=0;
    FOR(k, 2, N+1){
      ll n = N;
      while(n>=k){
        if(n%k==0){
          n /= k;
        }else{
          n -= k;
        }
      }
      if(n==1){
        debug(k);
        cnt++;
      }
    }
    p(cnt);

    
    return 0;
}