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.*;
import java.lang.Math;;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        ArrayList<Integer> natural = new ArrayList<>();
        ArrayList<Integer> minus = new ArrayList<>();
        for(int i=0;i<n;i++){
            int m = Integer.parseInt(sc.next());
            if(m<0){
                minus.add(m);
            }else{
                natural.add(m);
            }
        }
        sc.close();
        Collections.sort(natural);
        Collections.sort(minus, Collections.reverseOrder());
        List<int[]> q = new ArrayList<>();
        int ans;
        if(minus.size() > 0){
            int x = minus.get(0);
            for(int i=0;i<natural.size();i++){
                int y = natural.get(i);
                if(i < natural.size() - 1){
                    int[] query = {x,y};
                    q.add(query);
                    x -= y;
                }else{
                    int[] query = {y,x};
                    q.add(query);
                    x = y-x;
                }
            }
            for(int i=1;i<minus.size();i++){
                int y = minus.get(i);
                int[] query = {x,y};
                q.add(query);
                x -= y;
            }
            ans = x;
        }else{
            int x = natural.get(0);
            for(int i=0;i<natural.size();i++){
                int y = natural.get(i);
                if(i < natural.size() - 1){
                    int[] query = {x,y};
                    q.add(query);
                    x -= y;
                }else{
                    int[] query = {y,x};
                    q.add(query);
                    x = y-x;
                }
            }
            ans = x;
        }
        System.out.println(ans);
        for(int[] v : q){
            System.out.println(v[0] + " " + v[1]);
        }
    }
}