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;

const int INF = 1e9;
const int MOD = 1e9+7;
const ll LLINF = 1e18;

#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define ANS(n) cout << (n) << endl

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORR(i,a,b) for(int i=(a);i>=(b);i--)

#define ALL(obj) (obj).begin(),(obj).end()

#define pii pair<int,int>

#define pb(a) push_back(a)
#define mp make_pair

int main()
{
	int n;
	cin >> n;
	vector<int> p(n);
	FOR(i,0,n)cin >> p[i];
	sort(ALL(p));
	int ans = 0;
	FOR(i,0,n-1)ans += p[i];
	ans += p[n-1]/2;
    ANS(ans);
	return 0;
}