Problem:
Score : 500 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, written on a blackboard.

We will repeat the following operation N-1 times so that we have only one integer on the blackboard.

Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y.

Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.

Constraints

2 \leq N \leq 10^5

-10^4 \leq A_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value M of the final integer on the blackboard, and a sequence of operations x_i, y_i that maximizes the final integer, in the format below.

Here x_i and y_i represent the integers x and y chosen in the i-th operation, respectively.

If there are multiple sequences of operations that maximize the final integer, any of them will be accepted.

M
x_1 y_1
:
x_{N-1} y_{N-1}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in OCaml (4.02.3) (judge verdict: Accepted)

(* O(n) *)
let n = Scanf.scanf " %d" @@ (+) 0
let a_s = Array.init n @@ fun _ -> Scanf.scanf " %d" @@ (+) 0
let ps = Array.make_matrix 2 n 0
let pc = Array.make 2 0
let ms = [|[|max_int; min_int|]; [|max_int; min_int|]|]
let mis = Array.make_matrix 2 2 0
let ds = Array.make_matrix 2 n false
let xys = ref []
let f i a =
  let j = if a < 0 then 1 else 0 in
  ps.(j).(pc.(j)) <- a; pc.(j) <- pc.(j) + 1;
  if a < ms.(j).(0) then (ms.(j).(0) <- a; mis.(j).(0) <- i);
  if a >= ms.(j).(1) then (ms.(j).(1) <- a; mis.(j).(1) <- i)
let show_result () =
  Printf.printf "%d\n" ps.(0).(0);
  List.rev !xys |> List.iter (fun (x, y) -> Printf.printf "%d %d\n" x y)
let g () =
  for i = 2 to pc.(0) - 1 do
    let x, y = ps.(0).(1), ps.(0).(i) in
    xys := (x, y) :: !xys;
    ps.(0).(1) <- x - y
  done;
  let x, y = ps.(0).(0), ps.(0).(1) in
  xys := (x, y) :: !xys;
  show_result ()
let _ =
  Array.iteri f a_s;
  let i, j, k, l = if pc.(0) = 0 then 0, 1, 1, 0 else if pc.(1) = 0 then 1, 0, 0, 1 else -1, -1, -1, -1 in
  if i <> -1 then
    (if pc.(j) = 2 then
      let x, y = max ps.(j).(0) ps.(j).(1), min ps.(j).(0) ps.(j).(1) in
      Printf.printf "%d\n%d %d\n" (x - y) x y; exit 0
    else ();
    let x, y = ps.(j).(mis.(j).(k)), ps.(j).(mis.(j).(l)) in
    if x = y && i = 1 then (g (); exit 0);
    xys := (x, y) :: !xys;
    ds.(j).(mis.(j).(k)) <- true;
    ds.(j).(mis.(j).(l)) <- true;
    ps.(i).(0) <- x - y;
    pc.(i) <- 1)
  else ();
  let h (i0, j0) =
    for i = i0 to pc.(1 - j0) - 1 do
      if not ds.(1 - j0).(i) then
        let x, y = ps.(j0).(0), ps.(1 - j0).(i) in
        xys := (x, y) :: !xys;
        ps.(j0).(0) <- x - y;
        ds.(1 - j0).(i) <- true;
    done in
  List.iter h [1, 1; 0, 0]; show_result ()