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 lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for (int i = (begin), i##_end_ = (end); i < i##_end_; i++)
#define IFOR(i, begin, end) for (int i = (end)-1, i##_begin_ = (begin); i >= i##_begin_; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)

int main()
{
    lint n;cin >> n;
    vector<lint> v(n);
    lint ans = 0, x = 0;
    REP(i, n) { cin >> v[i];
        x = max(x, v[i]);
        ans += v[i];
    }

    cout << ans-x/2 <<"\n";

    return 0;
}