Problem:
Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1 \leq i \leq N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N-1 items cost their regular prices. What is the total amount he will pay?

Constraints

2 \leq N \leq 10

100 \leq p_i \leq 10000

p_i is an even number.

Input

Input is given from Standard Input in the following format:

N
p_1
p_2
:
p_N

Output

Print the total amount Mr. Takaha will pay.

Sample Input 1

3
4980
7980
6980

Sample Output 1

15950

The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen.

Note that outputs such as 15950.0 will be judged as Wrong Answer.

Sample Input 2

4
4320
4320
4320
4320

Sample Output 2

15120

Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

using namespace std ;

#define IOS ios::sync_with_stdio(0); cin.tie(0), cout.tie(0)
#define ll long long int 

#define fr first
#define se second

#define all(v) v.begin(),v.end() 
#define rall(v) v.rbegin(),v.rend()
#define sz(v) v.size() 

#define pb push_back
#define mp make_pair 

#define PI 3.141592653589793238



ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a*b / gcd(a,b) ;}


//bool valid(int x, int y){return x > 0 && x < n && y > 0 && y < m;}


int n;
int main(){
  cin >> n;
  vector<int>v(n);
  for(int i = 0 ; i < n ; i++)
    cin >> v[i];
  sort(all(v));
  int c = accumulate(v.begin(), v.end()-1, 0) + v[n-1]/2;
  cout << c;
  return 0;
}