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.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	static final PrintStream out = System.out;
	static Scanner sc = new Scanner(System.in);
	static final long Mod = 1000_000_007L;
	public static void main(String[] args) {
		Main main = new Main();
		main.solve();
		System.exit(0);
	}

	void solve() {
		int N = nextInt();
		int[] A = nextIntArray(N);
		Arrays.sort(A);
		int min = A[0],max = A[N-1];
		long sum = 0;
		int zero_i = Integer.MAX_VALUE;//0以上である最初のindex
		for(int i=0;i<N;i++) {
			if(zero_i == Integer.MAX_VALUE&&0<=A[i])zero_i=i;
			sum+=Math.abs(A[i]);
		}

		if(N == 2) {
			out.println(max-min);
			out.println(max+" "+min);
			return;
		}

		long res = 0;
		if(min<=0 && 0<=max) {//正数, 0, 負数のいずれか２つ以上存在
			res = sum;
		}else if(0 < min){//正数のみ
			res = sum - min - min;
		}else if(max < 0) {//負数のみ
			res = sum + max + max;
		}

		out.println(res);
		StringBuilder buf = new StringBuilder();

		sum = 0;
		if(min<=0 && 0<=max) {//正数, 0, 負数のいずれか２つ以上存在
			if(A[1]>=0) {
				sum = A[0]-A[N-1];
				buf.append(A[0] + " " +A[N-1]+System.getProperty("line.separator"));
			}else {
				sum = A[N-1]-A[0];
				buf.append(A[N-1] + " " +A[0]+System.getProperty("line.separator"));
			}
		}else if(0 < min){//正数のみ
			sum = A[0]-A[N-1];
			buf.append(A[0] + " " +A[N-1]+System.getProperty("line.separator"));
		}else if(0 > max) {//負数のみ
			sum = A[N-1]-A[0];
			buf.append(A[N-1] + " " +A[0]+System.getProperty("line.separator"));
		}
		for(int i=1;i<N-1&&i<zero_i;i++) {
			if(A[i+1]>=0) {
				if(i==N-2) {
					buf.append(sum+" "+A[i]+System.getProperty("line.separator"));
					sum = sum-A[i];
					continue;
				}
				buf.append(A[i]+" "+sum+System.getProperty("line.separator"));
				sum = A[i]-sum;
			}else if(A[i+1]<0) {
				buf.append(sum+" "+A[i]+System.getProperty("line.separator"));
				sum = sum-A[i];
			}
		}
		if(zero_i == 0)zero_i = 1;
		for(int i=zero_i;i<N-1;i++) {

			if(i == N-2) {
				buf.append(A[i]+" "+sum+System.getProperty("line.separator"));
				sum = A[i]-sum;
			}
			else {

				buf.append(sum + " " +A[i]+System.getProperty("line.separator"));
				sum = sum-A[i];
			}

		}

		out.println(buf);
	}
	private static int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)a[i] = nextInt();
		return a;
	}

	private static int nextInt() {
		return Integer.parseInt(next());
	}
	private static long nextLong() {
		return Long.parseLong(next());
	}
	private static String next() {
		return sc.next();
	}

}