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 Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
	n = readline() |> parseInt
	a = readline() |> split |> parseMap
	a = sort(a)
	m = 0
	h = div(n,2)
	for i in 1:n
		if i<=h
			m -= a[i]
		elseif i == h+1 && a[i] < 0
			m -=a[i]
		else
			m += a[i]
		end
	end
	println(m)
	k = 0
	if n%2 == 0
		k += 1
	end
	l = length(a)
	while l > 1
		if length(a)==2
			x = pop!(a)
			y = pop!(a)
			println(max(x,y), " ",min(x,y))
		elseif length(a)==3 && a[2] < 0
			x = shift!(a)
			println(a[2], " ", x)
			a[2] -= x
		elseif length(a)==3 && a[2]>0
			x = pop!(a)
			println(x," ",a[2])
			a[2] += x
		elseif k%2==1
			x = pop!(a)
			println(a[1], " ",x)
			a[1] -= x
		else
			x = shift!(a)
			println(x, " ", a[l-1])
			a[l-1] -= x
		end
		l -= 1
		k += 1
	end	
end

main()