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 double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 1, -1, 0};
int dy4[] = {1, 0, 0, -1};
const int MX = 1e6 + 123;
bitset<MX> is_prime;
vector<int> prime;
void primeGen(int n) {
  n += 100;
  for (int i = 3; i <= n; i += 2) is_prime[i] = 1;
  int sq = sqrt(n);
  for (int i = 3; i <= sq; i += 2) {
    if (is_prime[i] == 1) {
      for (int j = i * i; j <= n; j += (i + i)) {
        is_prime[j] = 0;
      }
    }
  }
  is_prime[2] = 1;
  prime.push_back(2);
  for (int i = 3; i <= n; i += 2) {
    if (is_prime[i] == 1) prime.push_back(i);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  long long p = sqrt(n);
  primeGen(p);
  for (long long i = 0; i < (int)prime.size(); i++) {
    if (n % prime[i] == 0) {
      cout << n / prime[i] << '\n';
      return 0;
    }
  }
  return 0;
}
