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 io() {}
const long long mod = 1e9 + 7;
const long long N = 1e6 + 7;
const long long maxn = 1e5 + 7;
long long n, m, a[N];
vector<long long> sp(maxn, 0);
void find_sp() {
  sp[1] = 1;
  for (long long i = 2; i <= maxn; i += 2) sp[i] = 2;
  for (long long i = 3; i * i <= maxn; i += 2) {
    if (!sp[i]) {
      sp[i] = i;
      for (long long j = i * i; j <= maxn; j += i) {
        if (!sp[j]) sp[j] = i;
      }
    }
  }
}
void __solve() {
  long long n;
  cin >> n;
  long long fp = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      fp = i;
      break;
    }
  }
  long long ans = 0;
  ans += n / fp;
  n %= fp;
  while (n > 0) {
    ans += n / sp[n];
    n %= sp[n];
  }
  cout << ans << '\n';
}
int32_t main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    __solve();
  }
}
