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;
using pii = pair<int, int>;
using ll = long long;
using llu = long long unsigned;
using ld = long double;
const ld EPS = 1e-9;
inline int cmp(ld first, ld second = 0, ld tol = EPS) {
  return (first <= second + tol) ? (first + tol < second) ? -1 : 0 : 1;
}
const int MOD = 1;
inline int mod(ll first, int m = MOD) { return (int)(((first % m) + m) % m); }
const int P = 1e5 + 100;
ll n;
bool not_prime[P];
vector<int> primes;
int main() {
  not_prime[0] = not_prime[1] = true;
  for (int i = 2; i * i < P; i++) {
    if (!not_prime[i])
      for (int j = i * i; j < P; j += i) not_prime[j] = true;
  }
  for (int i = 2; i < P; i++)
    if (!not_prime[i]) primes.push_back(i);
  scanf("%lld", &n);
  bool isPrime = true;
  int pos;
  for (int i = 0; i < ((int)(primes).size()); i++)
    if (n % primes[i] == 0) {
      isPrime = false;
      pos = (primes[i] - 3) / 2;
      break;
    }
  if (isPrime)
    printf("1\n");
  else if (n % 2 == 0 or n % 3 == 0)
    printf("%lld\n", n / 2);
  else
    printf("%lld\n", (n / 2) - pos);
  return 0;
}
