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.*;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		Integer[] A = new Integer[N];
		int sum=0;
		int all_plus=0;
		int all_minus=0;
		for (int i=0;i<N;i++) {
			A[i]=sc.nextInt();
			sum+=Math.abs(A[i]);
		}
		Arrays.sort(A);

		Queue<Integer> plus = new ArrayDeque<Integer>();
		Queue<Integer> minus = new ArrayDeque<Integer>();
		List<Integer> list = new LinkedList<Integer>();
		for (int i=0;i<N;i++) {
			if (A[i]>0) {
				plus.add(A[i]);
				all_plus=1;
			} else {
				list.add(A[i]);
				all_minus=1;
			}
		}
		Collections.reverse(list);
		for (int i=0;i<list.size();i++) {
			minus.add(list.get(i));
		}
		if (all_plus==0) {
			sum+=2*minus.peek();
		} else if (all_minus==0) {
			sum-=2*plus.peek();
		}
		System.out.println(sum);

		if (plus.isEmpty()) {
			int tmp=minus.remove();
			int tmp_2=minus.remove();
			System.out.print(tmp);
			System.out.print(" ");
			System.out.println(tmp_2);
			plus.add(tmp-tmp_2);
		}
		if (minus.isEmpty()) {
			int tmp=plus.remove();
			int tmp_2=plus.remove();
			System.out.print(tmp);
			System.out.print(" ");
			System.out.println(tmp_2);
			minus.add(tmp-tmp_2);
		}

		while (plus.size()>1) {
			int tmp=plus.remove();
			int tmp_2=minus.remove();
			System.out.print(tmp);
			System.out.print(" ");
			System.out.println(tmp_2);
			minus.add(tmp_2-tmp);
		}

		while (!minus.isEmpty()) {
			int tmp=plus.remove();
			int tmp_2=minus.remove();
			System.out.print(tmp);
			System.out.print(" ");
			System.out.println(tmp_2);
			plus.add(tmp-tmp_2);
		}
	}
}