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;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define REP(i,n) for(int i=0; i<(n); i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define ALL(n) begin(n),end(n)
#define IN(a, x, b) (a<=x && x<b)
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
template<class T> inline T CHMAX(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T CHMIN(T & a, const T b) { return a = (a > b) ? b : a; }

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);

int main()
{
    INIT;
    int n;
    cin >> n;
    int sum = 0;
    int ma = -1;
    int p;
    REP(i,n){
        cin >> p;
        sum += p;
        ma = max(ma, p); 
    }
    cout << sum - ma / 2 << endl;
    return 0;
}