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: Runtime Error)

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

public class Main {

	public static void main(String[] args) {

		try (Scanner sc = new Scanner(System.in)) {
			int n = sc.nextInt();
			List<Integer> a = new ArrayList<Integer>(); 
			for (int i = 0; i < n; i++) {
				a.add(sc.nextInt());
			}
			
			Collections.sort(a);
			int size = a.size();
			int temp = 0;

			if (size % 2 == 0) {
				while (a.size() > 1) {
					Collections.sort(a);
					System.out.println(a.get(a.size() - 1) + " " + a.get(0));
					temp = a.get(a.size() - 1) - a.get(0);
					a.remove(0);
					a.remove(a.size() - 1);
					a.add(temp);
					
					Collections.sort(a);
					System.out.println(a.get(0) + " " + a.get(a.size() - 1));
					temp = a.get(0) - a.get(a.size() - 1);
					a.remove(0);
					a.remove(a.size() - 1);
					a.add(temp);				
				}
				
			} else {
				Collections.sort(a);
				System.out.println(a.get(0) + " " + a.get(a.size() - 1));
				temp = a.get(0) - a.get(a.size() - 1);
				a.remove(0);
				a.remove(a.size() - 1);
				a.add(temp);
				
				Collections.sort(a);
				System.out.println(a.get(a.size() - 1) + " " + a.get(0));
				temp = a.get(a.size() - 1) - a.get(0);
				a.remove(0);
				a.remove(a.size() - 1);
				a.add(temp);				
			}
		}
	}
}