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>
#define ll long long
using namespace std;
vector<bool> prime(1e5 + 2,true);
find_prime_divisior(int n) {
	if(n==3) {
		return 3;
	}
	else {
		for(ll i=3;i<floor(sqrt(n))+1;i=i+2) {
			if(n%i==0 && prime[i]) {
				return i;
			}
		}
		return n;
	}
}
int main() {
	ll n;
	cin>>n;
	if(n%2==0) {
		cout<<n/2<<endl;
	}
	else {
		for(ll i=2;i<=1e5;i++) {
			if(prime[i]) {
				for(ll j=2*i;j<=1e5;j=j+i) {
					prime[j] = false;
				}
			}
		}
	ll divisior = find_prime_divisior(n);
	cout<<divisior<<endl;
	cout<<(n-divisior)/2 + 1<<endl;
	}
	return 0;
}