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 mod = 10e9 + 7;
long long power(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
vector<long long> pr;
void sieve() {
  long long n = 1e5 + 1;
  vector<bool> prime(n, true);
  int p = 2;
  while (p * p <= n) {
    if (prime[p]) {
      for (int i = 2 * p; i <= n; i += p) prime[i] = false;
    }
    p += 1;
  }
  for (int i = 2; i <= n; ++i) {
    if (prime[i]) pr.push_back(i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  sieve();
  long long n;
  cin >> n;
  long long ans = 0;
  while (n) {
    for (auto i : pr) {
      if (n % i == 0) {
        if (n % 2) {
          ans += 1;
          n -= i;
        } else {
          long long s = n / i;
          ans += s;
          n -= s * i;
          break;
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}
