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;
const double pi = 3.14159265359;
const long long INF = 1e12;
const long long MOD = 1e9 + 7;
const long long N = 1e5 + 5;
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  return i.second < j.second;
}
long long raise(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long n, ans = 0;
long long prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return i;
  }
  return n;
}
void solve() {
  cin >> n;
  cout << (1 + (n - prime(n)) / 2) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
