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;
int spf[100101];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 100101; i++) spf[i] = i;
  for (int i = 4; i < 100101; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 100101; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 100101; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
vector<int> getFactorization(int x) {
  vector<int> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
int main() {
  sieve();
  vector<int> pr;
  for (int i = 2; i < 100101; i++)
    if (spf[i] == i) pr.push_back(i);
  long long n;
  cin >> n;
  long long ans = 0;
  if (n % 2 == 0)
    ans = n / 2;
  else {
    for (int i = 0; i < pr.size(); i++)
      if (n % pr[i] == 0) {
        n -= pr[i];
        break;
      }
    ans = 1 + n / 2;
  }
  cout << ans << endl;
}
