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"
    #define LF '\n'
    #define ALL(x) x.begin(), x.end()
    #define iostreamBooster() do{cin.tie(nullptr);ios::sync_with_stdio(false);}while(0)
    using namespace std;
    typedef int64_t i64;
    typedef pair<int,int> pii;
    template<class A, class B>inline bool chmax(A &a, const B &b){return b>a ? a=b,1 : 0;}
    template<class A, class B>inline bool chmin(A &a, const B &b){return b<a ? a=b,1 : 0;}
    constexpr int INF = 0x3f3f3f3f;
    constexpr i64 LINF = 0x3f3f3f3f'3f3f3f3fLL;
     
     
    signed main()
    {
    int N;
    int a[105];
    int ans = 0;
     
    cin >> N;
    for (int i = 0; i < N; ++i) {
    cin >> a[i];
    ans += a[i];
    }
     
    sort(a, a+N);
    ans -= a[N-1];
    ans += a[N-1] / 2;
     
    cout << ans << endl;
     
     
    return 0;
    }