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;
using uint = unsigned int;
using lint = long long int;
using ulint = unsigned long long int;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T, class U> void assign(V<T>& v, int n, const U& a) { v.assign(n, a); }
template<class T, class... Args> void assign(V<T>& v, int n, const Args&... args) { v.resize(n); for (auto&& e : v) assign(e, args...); }


int main() {
  cin.tie(nullptr); ios_base::sync_with_stdio(false);
  int n; cin >> n;
  V<> p(n); for (int i = 0; i < n; ++i) cin >> p[i];
  sort(begin(p), end(p));
  int res = accumulate(begin(p), end(p), 0);
  res -= p[n - 1] >> 1;
  cout << res << '\n';
}