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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
const long long MOD = 1e9 + 7;
const long long mod = 998244353;
const int lim = 1e6;
int A[lim];
vector<int> primes;
void preprocess() {
  memset(A, 0, sizeof(A));
  for (int i = 2; i < lim; ++i) {
    if (A[i] == 0) {
      for (int j = 2 * i; j < lim; j = j + i) A[j] = 1;
    }
  }
  for (int i = 2; i < lim; ++i)
    if (A[i] == 0) primes.push_back(i);
}
int main() {
  preprocess();
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2ll << endl;
  } else {
    long long ans = n;
    for (int a : primes) {
      long long temp = 1ll * a;
      if (n % temp == 0) {
        ans = temp;
        break;
      }
    }
    long long ret = 1 + (n - ans) / 2;
    cout << ret << endl;
  }
  return 0;
}
