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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

integer N
integer,allocatable,dimension(:)::X
integer,allocatable,dimension(:)::ansx,ansy
integer ans
read*,N
allocate(X(N))
read*,X
call heapsort(N,X)

allocate(ansx(N-1),ansy(N-1))
if(mod(N,2)==1)then
  ansx(1)=x(1)
  ansy(1)=x(N)
  do i=2,N-1
    if(mod(i,2)==0)then
      ansx(i)=x(1+i/2)
    else
      ansx(i)=x(N-i/2)
    endif
      ansy(i)=ansx(i-1)-ansy(i-1)
  end do
else
  ansx(1)=x(N)
  ansy(1)=x(1)
  do i=2,N-1
    if(mod(i,2)==0)then
    ansx(i)=x(N-i/2)
  else
    ansx(i)=x(1+i/2)
  endif
    ansy(i)=ansx(i-1)-ansy(i-1)
  end do
endif

ans=ansx(N-1)-ansy(N-1)

print"(i0)",ans
do i=1,N-1
  print"(i0,A,i0)",ansx(i)," ",ansy(i)
end do
contains

subroutine heapsort(n,array)
  implicit none
!ここの入力は状況に応じて変更すること
  integer,intent(in) :: n
  integer,intent(inout) :: array(1:n)
  integer::i,k,j,l
  integer:: t

  l=n/2+1
  k=n
  do while(k /= 1)
     if(l > 1)then
        l=l-1
        t=array(L)
     else
        t=array(k)
        array(k)=array(1)
        k=k-1
        if(k == 1) then
           array(1)=t
           exit
        endif
     endif
     i=l
     j=l+l
     do while(j<=k)
        if(j < k)then
           if(array(j) < array(j+1))j=j+1
        endif
        if (t < array(j))then
           array(i)=array(j)
           i=j
           j=j+j
        else
           j=k+1
        endif
     enddo
     array(i)=t
  enddo
  return
end subroutine heapsort
end