Divisor Subtraction(CodeForces 1076B)

Description

You are given an integer number nn. 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 dd from nn and go to step 1.

Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer nn (2≤n≤10^10).

Output

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

Input

5

Output

1

Input

4

Output

2

Hint

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.

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\n",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;
}