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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        ArrayList<Integer> A = new ArrayList<>();
        for (int i=0; i<N; i++){
            A.add(sc.nextInt());
        }
        sc.close();

        int min,max,tmp;
        ArrayList<Integer> Z = new ArrayList<>();
        int i=0;
        while (A.size() != 2){
            Collections.sort(A);
            min = A.get(0);
            max = A.get(1);
            Z.add(min);
            Z.add(max);
            tmp = min - max;

//            System.out.println("A List");
//            for(int k=0; k<A.size(); k++){
//                System.out.print(A.get(k) + " ");
//            }
//            System.out.println();

            A.remove(0);
            A.remove(0);
            A.add(tmp);
            i = i+2;

//            System.out.println("A List");
//            for(int k=0; k<A.size(); k++){
//                System.out.print(A.get(k) + " ");
//            }
//            System.out.println();
//
        }
        Collections.sort(A);
        min = A.get(0);

//        System.out.println("A:"+A.size());
//        System.out.println("A List");
//        for(int k=0; k<A.size(); k++){
//            System.out.print(A.get(k) + " ");
//        }
//        System.out.println();


        max = A.get(A.size()-1);

        System.out.println(max-min);

//        System.out.println("Z:"+Z.size());
        for (int j=0; j<Z.size()-1; j=j+2){
            System.out.println(Z.get(j)+" "+Z.get(j+1));
        }
        System.out.println(max+" "+min);
    }
}