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;
const int maxn = 1e4 + 10;
char s[maxn];
long long check(long long x) {
  if (x == 1) return 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long i, j;
  long long n;
  int cnt = 0;
  cin >> n;
  long long w = n;
  long long ans = 1e18;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i) continue;
    if (check(i)) {
      while (n % i == 0) n /= i, cnt++;
      ans = min(ans, i);
    }
    if (n > i && check(n / i) && i * i != n) {
      long long k = n / i;
      ans = min(ans, k);
      while (n % k == 0) n /= k, cnt++;
    }
  }
  cout << w / ans << endl;
  return 0;
}
