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<stdio.h>
#include<iostream>
#include<iomanip>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include<string>
#include<math.h>
#include <numeric>

using namespace std;
typedef long long ll;

#define REP(i,n) for(int i=0;i<n;++i)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define pn(s) cout << #s << " " << s << endl

const ll mod = 1e9 + 7;
const ll inf = 1e18;

void printVector(vector<ll> A){
    cout << "[vector]" << endl;
    for(int i=0; i<A.size(); i++){
        cout << A[i] << " ";
    }
    cout << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    // input
    int N;
    cin >> N;

    vector<int> p;

    FOR(i, 0, N){
        int price;
        cin >> price;
        p.push_back(price);
    }    

    sort(ALL(p));

    int sum = accumulate(p.begin(), p.end()-1, 0) + p[N-1] / 2;

    p(sum);

    return 0;
}