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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <fstream>
#include <vector>
#include <queue>
#include <string>
#include <limits>
#include <algorithm>
#include <map>
#include <sstream>
#include <unordered_map>

using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
static ull tenq = 1000000000;
static ull mod = tenq + 7;

using namespace std;

int main()
{
  ll N;
  cin >> N;
  vector<ll> ps(N);
  for (auto i = 0; i < N; i++) cin >> ps[i];
  sort(ps.begin(), ps.end());
  ll res = 0;
  for (auto i = 0; i < N-1; i++) res += ps[i];
  res += ps[N-1] / 2;

  cout << res << endl;

  return 0;
}