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;
long long int MOD = (1e9) + 7;
void fastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
vector<bool> primes(200000, true);
bool isPrime(long long int x) {
  for (long long int i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
void sieve(long long int x) {
  for (long long int p = 2; p * p <= x; p++) {
    if (primes[p]) {
      for (long long int i = p * p; i <= x; i += p) {
        primes[i] = false;
      }
    }
  }
}
int main() {
  fastIO();
  long long int x;
  cin >> x;
  sieve(101010);
  if (x % 2 == 0) {
    cout << x / 2 << endl;
  } else {
    if (isPrime(x)) {
      cout << 1 << endl;
    } else {
      int ind = 2;
      while (x % ind != 0 || !primes[ind]) {
        ind++;
      }
      cout << x / ind << endl;
    }
  }
}
