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 REP1(i,n) for(int i=1, i##_len=(n); i<i##_len; ++i)
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define pow(x) x*x
#define ll long long
static const ll MAX = 1000000;
static const ll INFTY = 1e12;
static const ll MOD = 1000000007;

template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); };

struct edge {
    ll to, cost;
};

using Graph = vector<vector<int>>;
using _Graph = vector<vector<edge>>;


const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

// 深さ優先探索
vector<bool> seen;
vector<int> colors;

int H, W, N;
int sx, sy, gx, gy; 
vector<vector<char>> field;
vector<vector<int>> dist;
//vector<pair<int, int>> gx, gy;
deque<pair<int, int>> que;

typedef vector<vector<int> > Matrix;

const ll INF = 1e18;

int main() {

    int N;
    cin >> N;
    vector<int> p(N);
    REP(i, N) cin >> p[i];

    sort(p.begin(), p.end(), greater<int>());
    ll ans = 0;
    REP(i, N){
        if(i==0)p[i] = (int)(p[i]/2);
        ans += p[i];
    }

    cout << ans << endl;

}
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