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>
template <class Class>
void Print(Class instance) {
  std::cout << instance << std::endl;
}
void Print() { std::cout << std::endl; }
int main() {
  std::ios_base::sync_with_stdio(false);
  auto& input = std::cin;
  auto& output = std::cout;
  uint64_t n;
  input >> n;
  uint64_t answer = 0;
  while (n != 0) {
    if (n % 2 == 0) {
      answer += n / 2;
      break;
    }
    bool flag = false;
    for (uint64_t index = 3; index < std::sqrt(n) + 1; ++index) {
      if (n % index == 0) {
        ++answer;
        n -= index;
        flag = true;
        break;
      }
    }
    if (flag) {
      continue;
    }
    ++answer;
    break;
  }
  output << answer;
  return 0;
}
