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;
const int maxN = 1e5 + 7;
long long n;
bool P[maxN + 1];
vector<long long> S;
void sangNT() {
  memset(P, 0, sizeof(P));
  P[0] = P[1] = 1;
  S.clear();
  for (int i = 2; i < maxN; ++i) {
    if (P[i] == 0) {
      S.push_back(i);
      for (int j = 2; j <= maxN / i; ++j) {
        P[i * j] = 1;
      }
    }
  }
}
int main() {
  cin >> n;
  sangNT();
  int z = (int)S.size();
  for (int i = 0; i < z; ++i) {
    if (n % S[i] == 0) {
      cout << n / S[i] << endl;
      return 0;
    }
  }
  cout << 1;
  return 0;
}
