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 mod = 1e9 + 7;
long long N = 100311;
long long binpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1)
    return x % mod;
  else if (y % 2 == 0) {
    long long w = binpow(x, y / 2) % mod;
    return (w * w) % mod;
  } else
    return ((x % mod) * (binpow(x, y - 1) % mod)) % mod;
}
long long ans = 0;
void solve(long long n) {
  long long p = n;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      p = i;
      break;
    }
  }
  cout << n / p << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    solve(n);
  }
}
