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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

Var a,b,c:Array[1 .. 100000] of Longint;
    n,i,rem1,rem2,min,max,rem,count:Longint;
Begin
    Readln(n); min:=10000; max:=-10000; count:=0;
    For i:=1 to n do begin
        Read(a[i]);
        If a[i]<min then begin
            min:=a[i]; rem1:=i;
        End
        Else if a[i]>max then begin
            max:=a[i]; rem2:=i;
        End;
    End;
    rem:=min;
    For i:=1 to n do 
        If (i<>rem1) and (i<>rem2) then begin
            Inc(count);
            b[count]:=rem; c[count]:=a[i];
            rem:=rem-a[i];
        End;
    Writeln(max-rem);
    For i:=1 to count do writeln(b[count],' ',c[count]);
    Writeln(max,' ',rem);
End.