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 comp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
double EPSILON = numeric_limits<double>::epsilon();
bool AreSame(double a, double b) { return fabs(a - b) < EPSILON; }
long long primeFactors(long long n) {
  long long cnt = 0;
  while (n % 2 == 0) {
    cnt++;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      cnt++;
      n = n / i;
    }
  }
  if (n > 2) cnt++;
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long n;
  cin >> n;
  cout << primeFactors(n);
  return 0;
}
