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>

int main(void){

    int n;

    scanf("%d",&n);

    int p[n];

    for(int i=0;i<n;i++){

        scanf("%d",&p[i]);

    }

    for(int l=0;l<n-1;l++){

        for(int j=0;j<n-1;j++){

           if(p[j]<p[j+1]){

                int buf = p[j];

                p[j] = p[j+1];

                p[j+1] = buf;

            }

        }

    }

    int sum;

    p[0] = p[0]/2;

    for(int k=0;k<n;k++){

        sum += p[k];

    }

    printf("%d\n",sum);

    return 0;

}