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 long long MAXN = 1e5 + 5;
const long long mod = 998244353;
long long pw(long long b, long long r, long long md = mod) {
  b = b % mod;
  long long ans = 1;
  while (r) {
    if (r & 1) ans = (ans * b) % md;
    b = (b * b) % md;
    r >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    long long n;
    cin >> n;
    if (n % 2 == 0) {
      cout << (n / 2) << '\n';
      continue;
    }
    long long val = n;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        val -= i;
        break;
      }
    }
    if (val == n) {
      cout << 1 << '\n';
      continue;
    } else {
      cout << (val / 2) + 1 << '\n';
    }
  }
  return 0;
}
