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>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define ll long long
#define M 1000000007
#define sz(a) (ll)a.size()
#define pll pair<ll,ll>
#define rep(i,a,b) for(ll i=(ll)a;i<(ll)b;i++)
#define sep(i,a,b) for(ll i=(ll)a;i>=(ll)b;i--)
#define mll map<ll,ll>
#define vl vector<ll>
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(),a.end()
#define F first
#define S second
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll n,p;
    cin>>n;
    p=n;
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            p=i;
            break;
        }
    }
    cout<<n/p<<"\n";
}
