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 <iostream>
#include <string>
#include <vector>
#include <utility>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <map>
#include <cstdlib>
#include <queue>
#include <stack>
using namespace std;



int main(){
	int N;
	cin>>N;
	vector<int> a(N);
	for(int i=0;i<N;i++){
		cin>>a[i];
	}
	sort(a.begin(),a.end());
	int ans=0;
	for(int i=0;i<N;i++){
		if(i==N-1)ans+=a[i]/2;
		else ans+=a[i];
	}
	cout<<ans<<endl;
	return 0;
}