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)

open Printf
open Scanf

let (++) = Int64.add
let (--) = Int64.sub
let ( ** ) = Int64.mul
let (//) = Int64.div
let (%%) = Int64.rem
let landl = Int64.logand
let lorl = Int64.logor
let lnotl = Int64.lognot
let lxorl = Int64.logxor
let lsll = Int64.shift_left
let lsrl = Int64.shift_right_logical
let long = Int64.of_int
let succl = Int64.succ
let predl = Int64.pred

let read_int _ = scanf " %d " (fun x -> x)
let read_long _ = scanf " %Ld " (fun x -> x)
let read_str _ = scanf " %s " (fun x -> x)
let read_double _ = scanf " %f " (fun x -> x)

let m = 1_000_000_007L

let rec gcd a b = if b = 0L then a else gcd b (a %% b)

let rec mod_pow a = function
  | 0L -> 1L
  | b -> (if b %% 2L = 0L then 1L else a) ** mod_pow (a ** a %% m) (b//2L) %% m


let () =
  let n = read_int () in
  let p = Array.init n read_int in
  Array.sort compare p;
  p.(n-1) <- p.(n-1) / 2;
  let total = Array.fold_left ( + ) 0 p in
  printf "%d\n" total