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;
ifstream fi("test.txt");
ifstream fir("inp.txt");
ofstream fo("out.txt");
const long long MOD = 1e9 + 7;
const long long base = 269;
const int ooi = 1 << 30;
const int N = 1e5 + 5;
long long P(long long x, long long y) {
  long long ret = 1;
  while (y > 0) {
    if (y & 1) ret *= x;
    y /= 2;
    x = x * x;
  }
  return ret;
}
long long GCD(long long a, long long b) { return (b ? GCD(b, a % b) : a); }
long long n, m, q, k, pos, rs, check, t, T, s;
long long maxx = -1e9, minn = 1e9;
int p[N];
void init() {
  for (long long(i) = (2); (i) <= (N - 1); (i)++) {
    if (!p[i]) {
      p[i] = i;
      int j = 2;
      while (i * j <= N - 1) {
        p[i * j++] = i;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  init();
  while (n) {
    for (long long(i) = (2); (i) <= (N - 1); (i)++) {
      if (p[i] == i) {
        if (n % i == 0) {
          return cout << n / i, 0;
        }
      }
    }
  }
}
