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 prime[100001];
vector<int> pri;
void sieve() {
  memset(prime, 0, sizeof prime);
  for (int i = 2; i <= 100000; i++) {
    if (prime[i] == 1) continue;
    pri.push_back(i);
    for (int j = i + i; j <= 100000; j += i) {
      prime[j] = 1;
    }
  }
}
int main() {
  sieve();
  long long int a;
  cin >> a;
  bool find = 0;
  for (int i = 0; i < pri.size(); i++) {
    if (a % pri[i] == 0) {
      find = 1;
      printf("%lld\n", a / pri[i]);
      break;
    }
  }
  if (!find) printf("1\n");
}
