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>
     
    int setmax(int*,int);
     
    int main(void){
        int n;
        scanf("%d",&n);
        
        int i;
        int priceset[n];
        for(i=0;i<n;i++){
            scanf("%d",&priceset[i]);
        }
        
        int ans=0;
        int m = setmax(priceset,n);
        for(i=0;i<n;i++){
            if(priceset[i]==m){
                ans += m/2;
            }else{
                ans += priceset[i];
            }
        }
        printf("%d\n",ans);
    }
        
     
    int setmax(int data[],int n){
        int a =0;
        for(int i=0;i<n;i++){
            if(data[i]>a){
                a = data[i];
            }
        }
        return a;
    }