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.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);
	}
/*
-3 -3 2 4

 -3 4 => -7 -3 2
 -7 2 => -9 -3
 -9 -3 => 6

 -3 4 => -7 -3 2
 -3 2 => -7 5
 12

 -7 2 => 9 -3
 9 -3 => 12

3 3 2 4
 2 4 => 3 3 -2
 -2 3 => -5 3
 -5 3 => 8


-3 -3 -2 4
 4 -2 => 6 -3 -3
 6 -3 => 9 -3


 -3 -3 -2 -4
 -2 -3 => 1 -3 -4
 1 -3 => 4 -4
 4 -4 => 8


 2 4
 2 4 => 2

 -2 -4
 -2 -4 => 2

 3
 2 2 2
 2 2 => 0 2

3
 1 2 3
 -1 3






 */
	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 = -1;//0以上である最初のindex
		for(int i=0;i<N;i++) {
			if(zero_i == -1&&0<=A[i])zero_i=i;
			sum+=Math.abs(A[i]);
		}

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

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

		buf.append(A[0] + " " +A[N-1]+System.getProperty("line.separator"));

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

		out.println(buf);
	}
	void solve2() {
		int N = nextInt();
		int[] A = nextIntArray(N);
		Arrays.sort(A);
		int min = A[0],max = A[N-1];
		long sum = 0;
		for(int i=0;i<N;i++) {
			sum+=Math.abs(A[i]);
		}

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

		out.println(res);
	}
	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();
	}

}