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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <iterator>  // std::back_inserter()
#include <set>
#include <algorithm>  // std::copy()
#include <functional>  // std::greater<T>()
#include <utility>  // std::swap()
#include <numeric>  // accumulate(ALL(vec), 0)  0 は初期値
#include <cmath>
#include <climits>  // INT_MIN
#include <cctype>  // std::isdigit()
using namespace std;

#define ALL(obj) (obj).begin(), (obj).end()
#define REP(i, n) for (int i=0; i < (int)(n); i++)  // 0 ~ n-1
#define REPN(i, n) for (int i=1; i <= (int)(n); i++)  // 1 ~ n
#define MIN(vec) min_element(ALL((vec)))  // イテレータのため、値を取り出すときは * を先頭につける
#define MAX(vec) max_element(ALL((vec)))
#define IDX(vec, element_iter) distance((vec).begin(), element_iter)
#define SUM(vec) accumulate(ALL((vec)), 0)  // 0 は初期値
#define COPY(vec1, vec2) copy(ALL(vec1), back_inserter(vec2))  // vec1をvec2にコピーする vec2は空にしておく必要あり
typedef long long ll;

const int MOD = 1000000007;  // 1 000 000 007


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    int N;
    cin >> N;
    vector<int> vec(N);
    REP(i, N) cin >> vec[i];
    
    // calculation
    int idx = (int)IDX(vec, MAX(vec)), sum = 0;
    REP(i, N) {
        if (i == idx) {
            sum += vec[i]/2;
        } else {
            sum += vec[i];
        }
    }
    cout << sum << "\n";
    
    return 0;
}