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;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << a;
  err(++it, args...);
}
const int MAXN = 1e5 + 100;
const int oo = 1e9 + 100;
const int MOD = 1e9 + 7;
const long double EPS = 1e-9;
int main() {
  for (long long n, tc = 1; scanf("%lld", &n) == 1; ++tc) {
    ;
    long long p = -1;
    for (long long i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        p = i;
        break;
      }
    };
    long long ans = 0;
    if (n & 1) {
      if (p == -1) p = n;
      ans++;
      n -= p;
    }
    ans += n / 2;
    printf("%lld\n", ans);
  }
  return 0;
}
