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 OCaml (4.02.3) (judge verdict: Accepted)

let rec f maxN others lst =
  match lst with
  | [] -> (maxN, others)
  | x :: xs -> if x > maxN then f x (maxN :: others) xs else f maxN (x :: others) xs
let rec sum lst = match lst with
  | [] -> 0
  | x :: xs -> x + sum xs
let () =
  Scanf.scanf "%d\n"
    (fun n -> Array.init n (fun _ -> Scanf.scanf "%d\n" (fun x -> x)))
  |> Array.to_list |> f 0 []
  |> (fun (maxN, others) -> maxN/2 + sum others) |> print_int