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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    void run() {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        PriorityQueue<Integer> qPlus = new PriorityQueue<>();
        PriorityQueue<Integer> qMinus = new PriorityQueue<>(Collections.reverseOrder());

        int sum = 0;
        for (int i = 0; i < n; i++) {
            int a = sc.nextInt();
            if (a < 0) qMinus.add(a);
            else qPlus.add(a);
            sum += Math.abs(a);
        }

//        debug(qPlus);
//        debug(qMinus);

        if (qMinus.size() == 0) {
            int min = qPlus.peek();
            System.out.println(sum - min * 2);
            int tmp = qPlus.poll();
            for (int i = 0; i < n - 1; i++) {
                int a = qPlus.poll();
                if (qPlus.isEmpty()) {
                    System.out.println(a + " " + tmp);
                } else {
                    System.out.println(tmp + " " + a);
                    tmp = tmp - a;
                }
            }
        } else if (qMinus.size() == n) {
            int min = qMinus.peek();
            System.out.println(sum - Math.abs(min) * 2);
            int tmp = qMinus.poll();
            for (int i = 0; i < n - 1; i++) {
                int a = qMinus.poll();
                System.out.println(tmp + " " + a);
                tmp = tmp - a;
            }
        } else {
            System.out.println(sum);
            int qMinusSize = qMinus.size();
            int qPlusSize = qPlus.size();
            int tmp = qPlus.poll();
            for (int i = 0; i < qMinusSize - 1; i++) {
                int a = qMinus.poll();
                System.out.println(tmp + " " + a);
                tmp = tmp - a;
            }
            qPlus.add(tmp);
            int minusTmp = qMinus.poll();
            for (int i = 0; i < qPlusSize - 1; i++) {
                int a = qPlus.poll();
                System.out.println(minusTmp + " " + a);
                minusTmp = minusTmp - a;
            }
            System.out.println(qPlus.poll() + " " + minusTmp);
        }
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}