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;
long long minPrimeFactors(long long n) {
  long long minPrime = -1;
  if (n % 2 == 0) {
    minPrime = 2;
    return minPrime;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      int sign = 0;
      for (long long j = 2; j < i; j++) {
        if (i % j == 0) {
          sign = 1;
          break;
        }
      }
      if (sign == 0) {
        minPrime = i;
        return minPrime;
      }
    }
  }
  if (n > 2) minPrime = n;
  return minPrime;
}
int main() {
  long long n;
  cin >> n;
  cout << n / minPrimeFactors(n) << endl;
  return 0;
}
