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 int mod = 1e9 + 7;
vector<long long int> fact;
void factorize(long long n) {
  int count = 0;
  while (!(n % 2)) {
    n >>= 1;
    count++;
  }
  if (count) {
    fact.push_back(2);
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    count = 0;
    while (n % i == 0) {
      count++;
      n = n / i;
    }
    if (count) {
      fact.push_back(i);
    }
  }
  if (n > 2) fact.push_back(n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  bool prime = 1;
  factorize(n);
  for (int i = 0; i < fact.size(); i++) cout << fact[i] << endl;
  sort(fact.begin(), fact.end());
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) prime = 0;
  }
  if (prime) return cout << 1, 0;
  if (n % 2 == 0) return cout << n / 2, 0;
  return cout << (n - fact[0]) / 2, 0;
}
