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 p[100009];
vector<long long> prime;
void sieve() {
  p[2] = 0;
  prime.push_back(2);
  for (int i = 4; i <= 100009; i += 2) p[i] = 1;
  for (int i = 3; i <= 400; i += 2) {
    if (p[i] == 0) {
      for (int j = i * i; j <= 10009; j += 2 * i) {
        p[j] = 1;
      }
    }
  }
  for (int i = 3; i <= 100009; i += 2)
    if (p[i] == 0) prime.push_back(i);
}
int main() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
  } else {
    sieve();
    for (int i = 1; i < prime.size(); i++) {
      if (n % prime[i] == 0) {
        n -= prime[i];
        cout << n / 2 + 1;
        return 0;
      }
    }
    cout << 1;
  }
}
