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;
bool isPrime(long long n);
long long power(long long x, long long y);
long long gcd(long long a, long long b);
long long int mod = 1000000007;
void solve() {
  long long n{};
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return;
  }
  for (long long i = 2; i < n + 1; i++) {
    if (isPrime(i)) {
      if (n % i == 0) {
        cout << n / i;
        break;
      }
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long testcase = 1;
  while (testcase--) solve();
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
}
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2 || n == 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return ((power(x, y / 2) % mod) * (power(x, y / 2) % mod)) % mod;
  else
    return ((((x % mod) * ((power(x, y / 2) % mod))) % mod *
             (power(x, y / 2) % mod)) %
            mod) %
           mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
