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 num = sc.nextInt();
		long integer[] = new long[num];
		boolean plus[] = new boolean[num];
		for(int i = 0; i < num; i ++) {
			integer[i] = sc.nextLong();
		}
		Arrays.sort(integer);
		for(int i = 0; i < num; i ++) {
			plus[i] = i == 0 ? false : i == num - 1 ? true : integer[i] >= 0;
		}
		int count[] = {0, 0};
		int ans = 0;
		PriorityQueue<Long> lPlus = new PriorityQueue<>();
		PriorityQueue<Long> lMinus = new PriorityQueue<>();
		for(int i = 0; i < num; i ++) {
			count[plus[i] ? 1 : 0] ++;
			ans += integer[i] * (plus[i] ? 1 : -1);
			if(plus[i]) {
				lPlus.add(integer[i]);
			}else {
				lMinus.add(integer[i]);
			}
		}
		System.out.println(ans);


		for(int n = 1; n <= num - 1; n ++) {
			long tmp[] = {(long)lPlus.poll(), (long)lMinus.poll()};
			if(lPlus.size() <= lMinus.size()) {
				System.out.println(tmp[0]+" "+tmp[1]);
				lPlus.add(tmp[0] - tmp[1]);
			}else {
				System.out.println(tmp[1]+" "+tmp[0]);
				lMinus.add(tmp[1] - tmp[0]);
			}
		}
	}
}