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

int main(void){
        int N = 0;
        scanf("%d", &N);
        int l[N];
        for (int i = 0; i < N; i++) {
                        scanf("%d", &l[i]);
        }
        int tmp = -1;
        int sum = 0;
        for (int i = 0; i < N; i++) {
                        if (tmp < l[i]) tmp = l[i];sum += l[i];
        }
        printf("%d\n",sum - tmp / 2);
//      for (int i = 0; i < N; i++) {
//                      printf("%d\n",l[i]);
//      }
        return 0;
}