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 ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vs = vector<string>;

typedef pair<ll, ll> P;

#define bit(n) (1LL << (n))

//#define int long long

#define all(v) v.begin(), v.end()

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

template <class T>
inline void chmax(T& a, T b) {
  a = std::max(a, b);
}
template <class T>
inline void chmin(T& a, T b) {
  a = std::min(a, b);
}

#define mod (ll)(1e9 + 7)
#define INF LLONG_MAX

template <ll ModVal>
struct ModInt {
  ll x;

  ModInt(ll _x = 0) : x((_x % ModVal + ModVal) % ModVal) {
  }

  ModInt operator-() const {
    return ModInt(-x);
  }
  ModInt& operator+=(const ModInt a) {
    x += a.x;
    if (x >= ModVal)
      x -= ModVal;
    return *this;
  }
  ModInt& operator-=(const ModInt a) {
    x = x + ModVal - a.x;
    if (x >= ModVal)
      x -= ModVal;
    return *this;
  }
  ModInt& operator*=(const ModInt a) {
    x *= a.x;
    x %= ModVal;
    return *this;
  }

  ll ext_gcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    ll tmp = a / b;
    ll d = ext_gcd(b, a - b * tmp, y, x);
    y -= tmp * x;
    return d;
  }

  // 逆元
  ModInt inv(const ModInt a) {
    ll u, v;
    ext_gcd(a.x, ModVal, u, v);
    return ModInt(u);
  }

  ModInt& operator/=(const ModInt a) {
    return (*this) *= inv(a);
  }

  ModInt operator+(const ModInt a) const {
    ModInt retval(*this);
    return retval += a;
  }
  ModInt operator-(const ModInt a) const {
    ModInt retval(*this);
    return retval -= a;
  }
  ModInt operator*(const ModInt a) const {
    ModInt retval(*this);
    return retval *= a;
  }
  ModInt operator/(const ModInt a) const {
    ModInt retval(*this);
    return retval /= a;
  }

  ModInt pow(ll n) {
    ModInt ans(1);
    while (n) {
      if (n & 1)
        ans = ans * x;
      *this = (*this) * (*this);
      n = n >> 1;
    }
    return ans;
  }

  constexpr const ll& value() {
    return this->x;
  }
};

template <ll ModVal>
ostream& operator<<(ostream& os, const ModInt<ModVal>& a) {
  os << a.x;
  return os;
}

using mint = ModInt<mod>;

template <typename T>
class Combination {
 public:
  Combination(ll _max_n) : max_n(_max_n), factional(max_n + 1), inv(max_n + 1) {
    factional[0] = 1;
    inv[0] = 1;
    for (ll i = 0; i < max_n; i++) {
      factional[i + 1] = factional[i] * (i + 1); // n!(mod M)
      inv[i + 1] = inv[i] / (i + 1);             // k!^(M-2) (mod M)
    }
  }

  // nCk
  T choose(ll n, ll k) {
    if (n == 0 && k == 0)
      return 1;
    if (n < k || n < 0)
      return 0;
    T tmp = inv[n - k] * inv[k];
    return tmp * factional[n];
  }

  T permutation(ll n, ll k) {
    if (n - k < 0) {
      return 0;
    }
    return factional[n] / factional[n - k];
  }

 private:
  const ll max_n;
  std::vector<T> factional;
  std::vector<T> inv;
};

using Comb = Combination<mint>;

ll calc(ll n, ll k) {
  while (n != 1) {
    if (n % k) {
      n = n - k;
    } else {
      n = n / k;
    }
    if (n < 0) {
      return false;
    }
  }
  return true;
}

// prime factorization (素因数分解)
// key: value of prime factor
// value: number of prime factor
std::map<ll, ll> prime_factorization(ll n) {
  std::map<ll, ll> prime_factors;
  for (ll i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      prime_factors[i]++;
      n /= i;
    }
  }
  if (n != 1)
    prime_factors[n] = 1;
  return prime_factors;
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  ll n;
  cin >> n;
#if 0
  ll max_k = sqrt(n);

  ll ans = 0;
  for (ll k = 2; k <= max_k; k++) {
    ll tmp = n - k;
    if ((n - k) % (k * k) == 0) {
      ans++;
    }
  }

  if (max_k != n - 1) {
    ans++;
  }
  ans++;

  // cout << ans << endl;

  for (ll k = 2; k <= n; k++) {
    if (calc(n, k)) {
      cout << k << endl;
    }
  }
#endif

  auto factors = prime_factorization(n - 1);
  ll num_divisor = 1;
  for (auto it = factors.begin(); it != factors.end(); it++) {
    num_divisor *= it->second + 1;
  }

  cout << num_divisor + 1 << endl;

  return 0;
}