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 FOR(i,a,b) for(int i=(a);i<=(b);++i) 
#define RFOR(i,a,b) for(int i=(a);i>=(b);--i) 
#define REP(i,n) for(int i=0;i<(n);++i)
#define RREP(i,n) for(int i=(n);i>0;--i)
#define CI cin>> 
#define CO cout<<
#define EN <<endl 
#define all(x) (x).begin(),(x).end() 
#define asn ans
using namespace std;
template <typename T> istream& operator >> (istream& is, vector<T>& vec) { for (T& x : vec) is >> x; 	return is; }
using ll = long long;

int main() {
	int n; CI n;
	vector<int> p(n);  CI p;

	int ans=0;
	REP(i, n) {
		ans += p[i];
	}
	ans -= *max_element(p.begin(), p.end())/2;


	CO ans EN;
	return 0;
}