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 rep(i, n) for(int i = 0 ; i < (int)(n) ; i++)
typedef long long ll;
typedef long double Double;

#ifndef INPUTS_DIR
#define INPUTS_DIR "./"
#endif

using namespace std;

void solve(ll N, vector<ll> p) {
    sort(p.begin(), p.end());
    ll sum = 0;
    for (int i = 0; i < N - 1; i++) sum += p[i];
    cout << p.back() / 2 + sum << endl;
}


void main_(istream &cin) {
    ll N;
    cin >> N;
    vector<ll> p(N - 1 + 1);
    for (int i = 0; i <= N - 1; i++) {
        cin >> p[i];
    }
    solve(N, p);
}

int main() {
#ifdef INTELLIJ
    ifstream ifs(string(INPUTS_DIR) + "in_1.txt");
    if (ifs.is_open()) {
        main_(ifs);
        ifs.close();
    } else {
        cerr << "Error: no input." << endl;
    }
#else
    ios::sync_with_stdio(false);
    cin.tie(0);
    main_(cin);
#endif
}