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)

program main
  implicit none
  integer(8) i,N,Np,Nn,ip,in
  integer(8),allocatable :: A(:),pos(:),neg(:)
  read(*, *) N
  allocate(A(N))
  read(*, *) (A(i), i = 1,N)
  !正の数と負の数のカウントを行います。
  Np = 0
  Nn = 0
  do i = 1,N
    if (A(i) < 0) then
      Nn = Nn + 1
    else
      Np = Np + 1
    end if
  end do
  if (Nn == 0) then
    allocate(pos(N-1))
    allocate(neg(1))
    neg(1) = A(1)
    do i = 2,N
      if (A(i) <neg(1)) then
        pos(i-1) = neg(1)
        neg(1) = A(i)
      else
        pos(i-1) = A(i)
      end if
    end do
    write (*,*) sum(pos) -neg(1)
    do i = 1,N-2
      write(*, *) neg(1),pos(i)
      neg(1) = neg(1)-pos(i)
    end do
    write(*, *) pos(N-1),neg(1)

    
  else if (Np == 0) then
    allocate(pos(N))
    allocate(neg(N-1))
    pos(1) = A(1)
    do i = 2,N
      if (A(i) > pos(1)) then
        neg(i-1) = pos(1)
        pos(1) = A(i)
      else
        neg(i-1) = A(i)
      end if
    end do
    write (*,*) pos(1)-sum(pos) 
    do i = 1,N-1
      write(*, *) pos(1),neg(i)
      pos(1) = pos(1)-neg(i)
    end do

    
  else
    allocate(pos(Np))
    allocate(neg(Nn))
    ip = 1
    in = 1
    do i = 1,N
      if (A(i) < 0) then
        neg(in) = A(i)
        in = in + 1
      else
        pos(ip) = A(i)
        ip = ip + 1
      end if
    end do
    write(*, *) sum(pos)-sum(neg)
    if (Np > 1) then
      do i = 2,Np
        write(*, *) neg(1),pos(i)
        neg(1)=neg(1)-pos(i)
      end do
    end if
    if (Nn > 1) then
      do i = 2,Nn
        write(*, *) pos(1),neg(i)
        pos(1)=pos(1)-neg(i)
      end do
    end if
    write(*,*) pos(1),neg(1)
  end if
end program main