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 Kotlin (1.0.0) (judge verdict: Runtime Error)

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val A = readLine()!!.split(" ").map{it.toLong()}.sorted()

    val mn: List<Long>
    val ps: List<Long>

    if (A[0] > 0) {
        mn = A.take(1)
        ps = A.drop(1)
    } else if (A.last() < 0) {
        mn = A.dropLast(1)
        ps = A.takeLast(1)
    } else if (A.all { it == 0L }) {
        mn = A.take(1)
        ps = A.drop(1)
    } else {
        mn = A.filter { it < 0 }
        ps = A.filter { it >= 0 }
    }

    println(ps.sum() - mn.sum())

    var t = mn[0]

    for (a in ps.dropLast(1)) {
        println("$t $a")

        t -= a
    }

    println("${ps.last()} $t")

    t = ps.last() - t

    for (a in mn.drop(1)) {
        println("$t $a")
        t -= a
    }
}