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;
string s, s1;
long long int n, m, a, b, i, j, d, t, cs = 0, counT = 0, k, ans = 0, l = 0,
                                      temp = 0, sum = 0, Max, Min, num;
long long int arr[10000000];
bool Isprime[10000000];
void sv() {
  int t = sqrt(10000000);
  for (int i = 3; i <= t; i += 2) {
    if (!Isprime[i]) {
      for (int j = i * i; j < 10000000; j += i + i) Isprime[j] = true;
    }
  }
}
int main() {
  sv();
  cin >> n;
  while (n % 2 == 0 and n != 0) {
    counT = n / 2;
    n = 0;
  }
  if (n == 0) {
    return cout << counT, 0;
  }
  for (i = 3; i <= sqrt(n); i += 2) {
    if (!Isprime[i])
      if (n % i == 0) {
        counT++;
        n -= i;
        break;
      }
  }
  if (counT)
    cout << counT + n / 2;
  else
    cout << 1;
}
