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<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<complex>
#include<iostream>
#include<algorithm>
#define int long long
#define add_edge(u,v) nxt[++cnt]=head[u],head[u]=cnt,to[cnt]=v
#define open(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
char ch;bool fs;void re(int& x)
{
	while(ch=getchar(),ch<33);
	if(ch=='-')fs=1,x=0;else fs=0,x=ch-48;
	while(ch=getchar(),ch>33)x=x*10+ch-48;
	if(fs)x=-x;	
}
using namespace std;
int n,ans=1;
main()
{
	re(n);
	for(int i=2;i*i<=n;++i)
		if(n%i==0)
		{
			ans=n/i;
			break;
		}
	
	printf("%lld\n",ans);
}