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 <stdlib.h>
#include <math.h>
#include <limits.h>
#include <string.h>
#define ll long long int




int main()
{
   int N;

   scanf("%d",&N);
  int idx = 0;
  int array[N];
  int max = 0;
  for (int i = 0;i<N;i++)
  {
    scanf("%d",&array[i]);
    if (max<array[i])
       {
        max=array[i];
        idx = i;
       }
  }
  int sum = 0;

   for (int i =0;i<N;i++)
   {
       if (i==idx)
       sum+=array[i]/2;

       else if (i!=idx)
        sum+=array[i];
   }


   printf("%d",sum);

    return 0;
}