You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char c) { return '\'' + string(1, c) + '\''; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
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;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
string to_str(const int &n) {
  ostringstream stm;
  stm << n;
  return stm.str();
}
string to_str(const long long &n) {
  ostringstream stm;
  stm << n;
  return stm.str();
}
template <class T>
string tostring(T x, int len = 0) {
  stringstream ss;
  ss << x;
  string r = ss.str();
  if (int((r).size()) < len) r = string(len - int((r).size()), '0') + r;
  return r;
}
template <class T>
void convert(string x, T &r) {
  stringstream ss(x);
  ss >> r;
}
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long modinv(long long a) { return powmod(a, 1000000007 - 2); }
long long modinv(long long a, long long m) { return powmod(a, m - 2); }
long long nCkMp(long long n, long long k) {
  long long numerator = 1, denominator = 1;
  for (int i = 0; i < k; i++) numerator = (numerator * (n - i)) % 1000000007;
  for (int i = 1; i <= k; i++) denominator = (denominator * i) % 1000000007;
  return (numerator * modinv(denominator)) % 1000000007;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
int TT = 1, tt;
clock_t t1;
vector<bool> prime(100007, true);
bool isPrime(long long n) {
  if (n < 100007) return prime[n];
  if (n < 2) return false;
  if (n > 2 && n % 2 == 0) return false;
  for (long long i = 3; i * i <= n; i += 2)
    if (n % i == 0) return false;
  return true;
}
void sieve() {
  for (int p = 2; p * p < 100007; p++)
    if (prime[p])
      for (int i = p * 2; i < 100007; i += p) prime[i] = false;
}
void solve() {
  t1 = clock();
  long long n, res = 0;
  bool f;
  (cin >> (n));
  sieve();
  vector<int> primes;
  for (auto i = (2); i < (100007); i++)
    if (prime[i]) primes.push_back(i);
  while (n) {
    f = false;
    for (auto i = (0); i < (int((primes).size())); i++) {
      if (primes[i] > n) break;
      if (n % primes[i] == 0) {
        f = true;
        res += (n / primes[i]);
        n -= primes[i] * (n / primes[i]);
        404;
        break;
      }
    }
    if (!f) {
      res++;
      break;
    }
  }
  (cout << (res) << endl);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (tt = 0; tt < TT; tt++) solve();
  return EXIT_SUCCESS;
}
