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.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextLong();
		}
		sc.close();

		List<Long> list0 = new ArrayList<Long>();
		List<Long> list1 = new ArrayList<Long>();
		List<Long> list2 = new ArrayList<Long>();
		long sum = 0;
		long min = Integer.MAX_VALUE;
		long min1 = Integer.MAX_VALUE;
		long min2 = Integer.MIN_VALUE;
		for (int i = 0; i < n; i++) {
			if (a[i] == 0) {
				list0.add(a[i]);
			} else if (a[i] > 0) {
				list1.add(a[i]);
				min1 = Math.min(min1, a[i]);
			} else {
				list2.add(a[i]);
				min2 = Math.max(min2, a[i]);
			}
			long aa = Math.abs(a[i]);
			sum += aa;
			if (aa > 0) {
				min = Math.min(min, aa);
			}
		}

		long val = 0;
		if (list1.size() > 0 || list2.size() > 0) {
			if (list1.size() == 0) {
				if (list2.size() == 1) {
					System.out.println(sum);
				} else {
					System.out.println(sum - 2 * min);
				}
				val = min2;
				list2.remove(min2);
			} else if (list2.size() == 0) {
				if (list1.size() == 1) {
					System.out.println(sum);
				} else {
					System.out.println(sum - 2 * min);
				}
				val = min1;
				list1.remove(min1);
			} else {
				System.out.println(sum);
				if (list2.size() > 0) {
					val = min2;
					list2.remove(min2);
				}
			}
		} else {
			System.out.println(sum);
			list0.remove(list0.size() - 1);
		}

		for (int i = 0; i < list1.size() - 1; i++) {
			System.out.println(val + " " + list1.get(i));
			val -= list1.get(i);
		}
		if (list1.size() > 0) {
			long last = list1.get(list1.size() - 1);
			System.out.println(last + " " + val);
			val = last - val;
		}
		for (int i = 0; i < list2.size(); i++) {
			System.out.println(val + " " + list2.get(i));
			val -= list2.get(i);
		}
		for (int i = 0; i < list0.size(); i++) {
			System.out.println(val + " " + list0.get(i));
		}
	}
}