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 <iostream>
#include <vector>
#define maxn 100005
#define int long long
using namespace std;
bool npr[maxn];
vector<int> prm;
main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	for ( int i=2; i<maxn; ++i ) {
		if ( !npr[i] ) {
			for ( int j=i*i; j<maxn; j+=i ) {
				npr[i] = true;
				if ( i != 2 )
					prm.emplace_back(i);
			}
		}
	}
	long long n, ans = 0, x = -1;
	cin >> n;
	if ( n&1 ) {
		for ( int i : prm ) {
			if ( n%i == 0 ) {
				x = i;
				break;
			}
		}
		if ( x == -1 )	cout << 1 << '\n';
		else {
			if ( (n-x)&1 ) {
				cout << n/x << '\n';
			} else {
				cout << (n-x)/2+1 << '\n';
			}
		}
	} else {
		cout << n/2 << '\n';
	}
	return 0;
}
