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 std::pair;
using std::string;
using std::vector;
bool IsPrime(int64_t n) {
  if (n == 1) {
    return false;
  }
  if (n == 2) {
    return true;
  }
  for (int64_t i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  int64_t n;
  std::cin >> n;
  if (IsPrime(n)) {
    std::cout << 1;
    return 0;
  }
  int64_t i = 2;
  for (; !IsPrime(i) || n % i != 0; ++i) {
  }
  std::cout << n / i;
  return 0;
}
