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 successive_subtraction
  implicit none
  integer :: n, i
  integer(8) :: a(100000), m
  a = 0_8
  read(*,*) n
  read(*,*) a(1:n)
  call merge_sort(n,a(1:n))
  if (mod(n,2).eq.1) then
    m = a(n)
    do i = 1, n/2
      m = m+a(n-i)-a(i)
    end do
    write(*,'(i0)') m
    do i = 1, n/2
      write(*,'(i0,x,i0)') a(i), a(n-i)
    end do
    m = a(n)
    do i = 1, n/2
      write(*,'(i0,x,i0)') m, a(i)-a(n-i)
      m = m+a(n-i)-a(i)
    end do
  else
    m = a(n)-a(1)
    do i = 2, n/2
      m = m+a(n-i+1)-a(i)
    end do
    write(*,'(i0)') m
    write(*,'(i0)') a(n), a(1)
    do i = 2, n/2
      write(*,'(i0,x,i0)') a(i), a(n-i+1)
    end do
    m = a(n)-a(1)
    do i = 2, n/2
      write(*,'(i0,x,i0)') m, a(i)-a(n-i+1)
      m = m+a(n-i+1)-a(i)
    end do
  end if
  stop
contains
  subroutine merge_sort(n,c)
    implicit none
    integer, intent(in) :: n
    integer(8), intent(inout) :: c(n)
    integer :: m, l, i, u
    m = n
    l = 1
    do while (m.gt.1)
      do i = 1, m/2
        u = min(2*i*l,n)
        call merge(c(2*(i-1)*l+1:(2*i-1)*l),c((2*i-1)*l+1:u))
      end do
      l = 2*l
      m = (m+1)/2
    end do
    return
  end subroutine merge_sort
  subroutine merge(c1,c2)
    implicit none
    integer(8), intent(inout) :: c1(:), c2(:)
    integer(8) :: c(size(c1)+size(c2))
    integer :: i1, i2, n1, n2
    i1 = 1
    i2 = 1
    n1 = size(c1)
    n2 = size(c2)
    do while (i1.le.n1.and.i2.le.n2)
      if (c1(i1).le.c2(i2)) then
        c(i1+i2-1) = c1(i1)
        i1 = i1+1
      else
        c(i1+i2-1) = c2(i2)
        i2 = i2+1
       end if
    end do
    if (i1.le.n1) then
      c(i1+i2-1:n1+n2) = c1(i1:n1)
    else if (i2.le.n2) then
      c(i1+i2-1:n1+n2) = c2(i2:n2)
    end if
    c1 = c(1:n1)
    c2 = c(n1+1:n1+n2)
    return
  end subroutine merge
end program successive_subtraction