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 fi first
#define se second
#define p_b push_back
#define pll pair<ll,ll>
#define m_p make_pair
#define all(x) x.begin(),x.end()
#define ld long double
#define sset ordered_set

using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
const ll MAXN = 1123456;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <typename T>
T sqr(T x){
    return x * x;
}

template <typename T>
void vout(T s){
    cout << s << endl;
    exit(0);
}

ll bp(ll a,ll n){
    ll res = 1;
    while(n){
        if(n % 2)res *= a;
        a *= a;
        n >>= 1;
    }
    return res;
}



int main(){
    ios_base :: sync_with_stdio(0);
    cin.tie(0);
    //freopen("numbers.in","r",stdin);
    //freopen("numbers.out","w",stdout);


    ll n;
    cin >> n;
    for(int i = 2; i * i <= n; i++)if(n % i == 0){
        vout(n / i);
    }
    vout(1);
    cerr << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec" << endl;
    return 0;
}
