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 N = 200005;
bool col2(pair<long long, long long> p1, pair<long long, long long> p2) {
  if (p1.second == p2.second) return p1.first > p2.first;
  return p1.second > p2.second;
}
struct comp {
  bool operator()(long long const &x, long long const &y) { return x > y; }
};
vector<pair<long long, long long>> prime;
void f(long long n) {
  long long count = 0;
  while (!(n % 2)) {
    n >>= 1;
    count++;
  }
  if (count) prime.push_back(make_pair(2, count));
  for (long long i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n = n / i;
    }
    if (count) prime.push_back(make_pair(i, count));
  }
  if (n > 2) prime.push_back(make_pair(n, 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long n;
  cin >> n;
  f(n);
  if (prime[0].first % 2 == 0) {
    cout << (n / prime[0].first);
  } else {
    cout << 1 + ((n - prime[0].first) / 2);
  }
}
