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;
vector<int> v;
bool prime[100005];
void findPrime(int n) {
  for (int i = 4; i <= n; i += 2) {
    prime[i] = true;
  }
  for (int i = 3; i * i <= n; i += 2) {
    if (!prime[i]) {
      for (int j = 2; i * j <= n; j++) {
        prime[i * j] = true;
      }
    }
  }
  v.push_back(2);
  for (int i = 3; i <= n; i = i + 2) {
    if (!prime[i]) {
      v.push_back(i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  findPrime(100000);
  long long int a, b, c, d, i, cnt = 0, chk = 0;
  cin >> a;
  for (i = 0; i < v.size(); i++) {
    if (a % v[i] == 0) {
      chk = 1;
      cout << a / v[i];
      return 0;
    }
  }
  if (chk == 0) {
    cout << 1;
  }
  return 0;
}
