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 <algorithm>
#include <vector>
using namespace std;
int main(void)
{
    int n = 0;
    int ans = 0;
    vector<int> p(n);
    cin >> n;
    for (size_t i = 0; i < n; i++)
    {
        int tmp;
        cin >> tmp;
        p.push_back(tmp);
    }
    sort(p.begin(), p.end());
    for (size_t i = 0; i < n - 1; i++)
    {
        ans += p[i];
    }
    ans += p[n - 1] / 2;
    cout << ans << endl;
    return 0;
}