Divisor Subtraction

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 n 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.

Sample Input

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.

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\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=;i*i<=n;i++)
     {
         if(n%i==)
         {
             cout<<+(n-i)/<<endl;
             return ;
         }
     }
     cout<<""<<endl;
     return ;
 }