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: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        sc.close();
        Arrays.sort(a);

        int minpos = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] > 0) {
            }
            if (a[i] < 0) {
                minpos = i;
            }
        }
        if (n == 2) {

            System.out.println(a[n - 1] - a[0]);
            System.out.println(a[n - 1] + " " + a[0]);
            return;
        }
        long sum = 0;
        if (minpos == n - 1) {
            for (int i = 0; i < n - 1; i++) {
                sum += a[i];
            }
            sum = a[n - 1] - sum;
            System.out.println(sum);
            long now = a[n - 1];
            for (int i = 0; i < n - 2; i++) {
                System.out.println(now + " " + a[i]);
                now += a[i];
            }
            System.out.println(a[n - 1] + " " + now);
            now = a[n - 1] - now;
            return;

        }
        long ans = a[0];
        for (int i = minpos + 1; i < n - 1; i++) {
            // 正の数を負にかえる // a[n - 1]以外負の数
            ans -= a[i];
        }
        ans = a[n - 1] - ans;
        for (int i = 1; i <= minpos; i++) {
            if (minpos == 0) {
                break;
            }
            ans -= a[i];
        }
        // ans = a[n - 1] - ans;
        System.out.println(ans);

        long now = a[0];
        for (int i = minpos + 1; i < n - 1; i++) {
            System.out.println(now + " " + a[i]);
            ans -= a[i];
            now -= a[i];
        }
        System.out.println(a[n - 1] + " " + now);
        now = a[n - 1] - now;
        for (int i = 1; i <= minpos; i++) {
            if (minpos == 0) {
                break;
            }
            System.out.println(now + " " + a[i]);
            ans -= a[i];
            now -= a[i];
        }

    }
}