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 <string.h>
#include <math.h>
 
int main()
{
  int  a, b, c, d, X, Y;
//  double  c;
//  long long  a, b, c, d, X, Y;
  int D, G;
  //char s[101][101]={};
  char sv[101]={};
  char ss[101]={};
//  char s_temp[101]={};
  char t[12]={};
//  char s[10000000000000000001]={};
  char s[2][100001]={};
  int len=0;
//  int N, M;
  int A[11]={};
//  long long  A[101]={};
//  long int A[1001]={};
  long long  B[101]={};
  long long  C[101]={};
  int i,j,k;
  int flag=0;
  int tmp=0;
  int min=1000;
  int cnt=0;
  double ans=1000;
  int max=-1;

  int  Cen_X=0;
  int  Cen_Y=0;
  long long  H=0;
  
  //  int ans=0;
  int str_len=0;
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
  int ans_o[1001]={};
  double N[1001]={};
  int ans_e[100001]={};
//  scanf("%lld %lld", &a, &b);
  scanf("%d", &a);
  b=0;
  for(i=0;i<a;i++){
    scanf("%d", &c);
    if(max<c) max=c;
    b=b+c;
  }
  b=b-(max/2);

  
  printf("%d\n", b);
  
return 0;
}