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: Runtime Error)

    #include<stdio.h>
    #include<string.h>
    int main(void){
    int A,B,C,D,E,N,M,x,y,i,j,cou=0,cou2,v[1000000],w[1000000],min_i,en=0.0;
        char S[100][100],W[100],s[10],X,Y;
    scanf("%d",&A);
    for(i=0;i<A;i++){
        scanf("%d",v[i]);
    }
    min_i=0;
    for(i=1;i<A;i++){
        if(v[i]>v[min_i]){
            min_i=i;
        }
    }
    v[min_i]=v[min_i]/2;
    for(i=0;i<A;i++){
        cou+=v[i];
    }
    printf("%d",cou);
    return 0;
    }