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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
const ll oo = 0x3f3f3f3f;
const ll MOD = 1000000007;
const ll N = 1e6;
bool isP[N];
void build() {
  fill(isP, isP + N, 1);
  isP[0] = isP[1] = 0;
  for (ll i = 2; i * i <= N; i++)
    if (isP[i])
      for (ll j = i * i; j <= N; j += i) isP[j] = false;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  build();
  ll x;
  cin >> x;
  ll ans = 0;
  for (ll i = 2; i < N; i++) {
    if (!isP[i]) continue;
    while (x % i == 0) x /= i, ans++;
  }
  if (x != 1) ans++;
  cout << ans << '\n';
  return 0;
}
