Educational Codeforces Round 54 (Rated for Div. 2) B. Divisor Subtraction

传送：B. Divisor Subtraction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer number n n . The following algorithm is applied to it:

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

Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n n (2≤n≤10 10  2≤n≤1010 ).

Output

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

Examples

Input

Copy

5

Output

Copy

1

Input

Copy

4

Output

Copy

2

Note

In the first example 5 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0 0 .

In the second example 2 2 is the smallest prime divisor at both steps.

题解：一个数减去最小的质数之后的最小质数必然是2。知道这个，这个题就很明了了。

分为两种： 一种是像5，3这种的质数，直接减自己，结果就是1。

                    另一种是非质数，找见第一个质数之后，用减去质数的数字/2就是还能继续减的次数然后还要加1，因为减去第一个质数的时候有一次步骤。

如 n=21 变化如下：

 21—18—16—14—12—10—8—6—4—2—0

结果是10

#include <iostream>

using namespace std;
typedef long long ll;
int main(){
    ll n; cin>>n;
    ll x;
    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;
}