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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

#define llong long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12

int main(void)
{
	//変数の宣言
	int n,m;
	int p[10010];
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	llong int ans=0;
	int count=0;
	int temp,temp1,temp2;
	llong int max,min;
	int len;
	int sum=0;
	//データの読み込み
	scanf("%d",&n);
	
	for(i=0;i<n;i++){
		scanf("%d",&p[i]);
	}
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	for(i=0;i<n;i++){
		if(i==0){
			max=p[0];
		}else{
			max=max(max,p[i]);
		}
		sum=sum+p[i];
	}
	
//	printf("計算部分終了\n");
	//出力
	
	printf("%d",sum-max/2);
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}