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;
vector<long long> prime;
void seive() {
  long long n = 100000;
  bool flag[100001];
  memset(flag, 0, sizeof(flag));
  flag[0] = true;
  flag[1] = true;
  for (long long i = 2; i * i <= n; i++) {
    if (flag[i] == false) {
      prime.push_back(i);
      for (long long j = i + i; j <= n; j += i) {
        flag[j] = true;
      }
    }
  }
}
int main() {
  seive();
  long long n;
  cin >> n;
  long long ans = 1;
  long long i = 0;
  while (i < prime.size() && prime[i] * prime[i] <= n) {
    if (n % prime[i] == 0) {
      n -= prime[i];
      ans += n / 2;
      break;
    }
    i++;
  }
  cout << ans << endl;
  return 0;
}
