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;
bool is_prime(long long x) {
  int flag = 1;
  for (long long i = 2; i <= x / 2; i++) {
    if (x % i == 0) {
      flag = 0;
      break;
    }
  }
  if (flag == 1) {
    return true;
  } else if (flag == 0) {
    return true;
  }
}
long long find_prime_factors(long long n) {
  long long val = n;
  std::vector<long long> result;
  for (long long i = 2; i <= n; i++) {
    if (is_prime(i)) {
      while (val % i == 0 && val != 1) {
        result.push_back(i);
        val = val / i;
      }
      if (val == 1) {
        break;
      }
    }
  }
  return result.size();
}
int main() {
  long long x;
  cin >> x;
  cout << find_prime_factors(x);
  return 0;
}
