Divisor Subtraction

Description

You are given an integer number $\mathrm{nn.\; The\; following\; algorithm\; is\; applied\; to\; it:}$

1. if $\mathrm{n=0,\; then\; end\; algorithm;}$
2. find the smallest prime divisor $\mathrm{d of n;}$
3. subtract $\mathrm{dd from n and\; go\; to\; step 1.}$

Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer $\mathrm{nn (2\le n\le 10^10).}$

Output

Print a single integer — the number of subtractions the algorithm will make.

Sample Input

Input

5

Output

1

Input

4

Output

2

Hint

In the first example $\mathrm{5 is\; the\; smallest\; prime\; divisor,\; thus\; it\; gets\; subtracted\; right\; away\; to\; make\; a 0.}$

In the second example $\mathrm{2 is\; the\; smallest\; prime\; divisor\; at\; both\; steps.}$

题解：给你一个数n，找它的最小素因子，每回减去，直到n=0，素因子均为奇数（除了2），奇-奇=偶，例如15，标准答案应该为7而不是5，在1e5内没有素因子时，答案为1。

代码如下：

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<queue>
#include<map>
#include<set>
#define memset(x,y) memset(x,y,sizeof(x))
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define debug(x) cout<<x<<" "<<endl
#define rson i << 1 | 1,m + 1,r
#define e  2.718281828459045
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define read(x) scanf("%d",&x)
#define put(x) printf("%d
",x)
#define lowbit(x) (x&(-x))
#define lson i << 1,l,m
#define INF 0x3f3f3f3f
#define ll long long
#define mod 1001113
#define N 100000000
#define PI acos(-1)
#define eps 1.0e-6
#define maxn 27
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
//const int maxn=;
using namespace std;



int main()
{
    ll n;
    cin>>n;
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            cout<<1+(n-i)/2<<endl;
            return 0;
        }
    }
    cout<<"1"<<endl;
    return 0;
}