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 long long maxN = 1e5 + 10 + 10;
bitset<maxN> bs;
vector<long long> p;
void _sieve(long long n) {
  bs.set();
  bs[0] = bs[1] = 0;
  for (long long i = 2; i <= n; i++) {
    if (bs[i]) {
      for (long long j = i * i; j <= n; j++) bs[j] = 0;
      p.push_back(i);
    }
  }
}
int32_t main() {
  long long n;
  cin >> n;
  if (n & 1) {
    long long spd = n;
    _sieve(maxN - 10);
    for (long long i = 0; i < (long long)p.size() && p[i] * p[i] <= n; i++) {
      if (n % p[i] == 0) {
        spd = p[i];
        break;
      }
    }
    n -= spd;
    cout << 1 + (n >> 1);
  } else
    cout << (n >> 1);
}
