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>
#pragma comment(linker, "/stack:336777216")
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using ui = unsigned int;
template <typename T>
using less_priority_queue = priority_queue<T, vector<T>, greater<T>>;
ll const mod = 1e9 + 7;
auto mt = mt19937(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  ll n;
  cin >> n;
  if (n % 2 == 0) {
    cout << (n / 2) << '\n';
  } else {
    set<ll> divs;
    ll n2 = n;
    divs.insert(n);
    for (ll i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        divs.insert(i);
        while (n % i == 0) n /= i;
      }
    }
    n2 -= *divs.begin();
    cout << (n2 / 2) + 1 << '\n';
  }
  return 0;
}
