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 log n) *)
let n = Scanf.scanf " %d" @@ (+) 0
let a_s = Array.init n @@ fun _ -> Scanf.scanf " %d" @@ (+) 0
let p = ref @@ -1
let xys = ref []
let _ =
  Array.sort (-) a_s;
  let i = ref 0 in while !p < 0 && !i < n do if a_s.(!i) >= 0 then p := !i; incr i done;
  if !p = -1 then p := n - 1
  else if !p = 0 then p := 1;
  for i = !p to n - 2 do
    xys := (a_s.(0), a_s.(i)) :: !xys;
    a_s.(0) <- a_s.(0) - a_s.(i)
  done;
  for i = 0 to !p - 1 do
    xys := (a_s.(n - 1), a_s.(i)) :: !xys;
    a_s.(n - 1) <- a_s.(n - 1) - a_s.(i)
  done;
  Printf.printf "%d\n" a_s.(n - 1);
  List.rev !xys |> List.iter @@ fun (x, y) -> Printf.printf "%d %d\n" x y