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.*;
 
public class Main{
  public static void main(String[]args){
    Scanner sc =  new Scanner(System.in);
    int N = Integer.parseInt(sc.nextLine());
    String[] S = sc.nextLine().split(" ");
    List<Integer> list1 = new ArrayList<>();
    List<Integer> list2 = new ArrayList<>();
    
    for(int i = 0; i < N; i++){
      int n = Integer.parseInt(S[i]);
      if(n > 0){
        list1.add(n);
      }else{
        list2.add(n);
      }
    }
    
    Collections.sort(list1);
    Collections.sort(list2);
    int ans = 0;
    int[][] val = new int[N-1][2];
    for(int i = 0; i < N-1; i++){
      if(list2.size() > 1){
        int a = list2.get(0);
        list2.remove(0);
        int b = list2.get(0);
        list2.remove(0);
        list2.add(a-b);
        val[i][0] = a;
        val[i][1] = b;
      }else if(list2.size() == 1 && list1.size() == 1){
        ans = list1.get(0) - list2.get(0);
        val[i][0] = list1.get(0);
        val[i][1] = list2.get(0);
      }else if(list2.size() == 0){
        int a = list1.get(0);
        list1.remove(0);
        int b = list1.get(0);
        list1.remove(0);
        val[i][0] = a;
        val[i][1] = b;
        if(a-b > 0){
          list1.add(a-b);
        }else{
          list2.add(a-b);
        }
      }else{
        int a = list2.get(0);
        list2.remove(0);
        int b = list1.get(0);
        list1.remove(0);
        list2.add(a-b);
        val[i][0] = a;
        val[i][1] = b;
      }
    }
    System.out.println(ans);
    for(int i = 0; i < N-1; i++){
      System.out.print(val[i][0]);
      System.out.print(" ");
      System.out.println(val[i][1]);
    }
  }
}