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: Wrong Answer)

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

int sumof(int p[],int N)
{
  int max, minelse, sum;
  for(int i=0; i<N; i++){
    if(p[i] > max){
      	max = p[i] - (p[i]*0.5);
    }else{
     minelse += p[i];
    }
  }
  sum = max + minelse;
  return sum;
}

int main(void)
{
  int *n;
  int N;
  
  n = calloc(N, sizeof(int));
  
  for(int i=0; i<N; i++){
   scanf("%d", &n[i]);
  }
  printf("%d", sumof(n,N));
  free(n);
  return 0;
  
}