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 n, t, b[2000005], mx, mn = 1e9, ans, l, k = 1e9;
int main() {
  cin >> n;
  if (n <= 10000) {
    for (long long i = 2; i <= sqrt(n); i++)
      if (b[i] == 0) {
        for (long long j = i + i; j <= n; j += i) b[j] = 1;
      }
    for (long long i = 2; i <= n; i++) {
      if (b[i] == 0)
        if (n % i == 0) mx = i, mn = min(mn, mx);
    }
    return cout << (n - mn) / 2 + 1, 0;
  } else {
    for (long long i = 2; i * i <= sqrt(n); i++)
      if (b[i] == 0) {
        for (long long j = i + i; j * j <= n; j += i) b[j] = 1;
      }
    for (long long i = 2; i * i <= sqrt(n); i++) {
      if (b[i] == 0)
        if (n % i == 0) l = i, k = min(k, l);
    }
    return cout << (n - k) / 2 + 1, 0;
  }
}
