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 {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[] a = new int[n];
    int minus = 0;
    int index = 0;
    int z = 100000;
    int sum = 0;
    int m1 = 0;
    ArrayList<Integer> listp = new ArrayList<Integer>();
    ArrayList<Integer> listm = new ArrayList<Integer>();
    int[] mm = new int[n];
    for(int i = 0; i < n; i++) {
      a[i] = sc.nextInt();
      sum += a[i];
      if(a[i] < 0) {
        minus++;
        m1 += a[i];
        listm.add(i);
        mm[i] = 1;
      } else {
        listp.add(i);
      }
      if(Math.abs(a[i]) < z) {
        index = i;
        z = Math.abs(a[i]);
      }
    }
    if(minus == 0) {
      System.out.println(sum - 2 * a[index]);
      if(n == 2) {
        System.out.println(a[1 - index] + " " + a[index]);
      } else {
        int t = a[index];
        int p = 0;
        for(int i = 0; i < n; i++) {
          if(i == index) {
            
          } else {
            if(p == n - 1) {
              
            } else if(p == n - 2) {
              System.out.println(a[i] + " " + t);
            } else {
              System.out.println(t + " " + a[i]);
              t -= a[i];
            }
            p++;
          }
        }
      }
    } else if(minus == n) {
      System.out.println(2 * a[index] - sum);
      int t = a[index];
      for(int i = 0; i < n; i++) {
        if(i == index) {
        } else {
          System.out.println(t + " " + a[i]);
          t -= a[i];
        }
      }
    } else {
      System.out.println(sum - 2 * m1);
      int t = a[listm.get(0)];
      for(int i = 0; i < listp.size(); i++) {
        if(i == listp.size() - 1) {
          System.out.println(a[listp.get(i)] + " " + t);
          t = a[listp.get(i)] - t;
        } else {
          System.out.println(t + " " + a[listp.get(i)]);
          t -= a[listp.get(i)];          
        }
      }
      for(int i = 1; i < listm.size(); i++) {
        System.out.println(t + " " + a[listm.get(i)]);
        t -= a[listm.get(i)];
      }
    }
  }
}