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.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		List<Long> list_p = new LinkedList<>();
		List<Long> list_n = new LinkedList<>();
		long ans = 0;
		for(int i=0;i<n;i++) {
			long tmp = in.nextInt();
			if(tmp>=0) {
				list_p.add(tmp);
				ans += tmp;
			}else {
				list_n.add(tmp);
				ans -= tmp;
			}
		}
		if(list_p.size()>1) list_p.sort((e1,e2)->e1>e2?-1:1);
		if(list_n.size()>1) list_n.sort((e1,e2)->e1<e2?-1:1);
		if(list_p.size()==0) ans += 2*list_n.get(list_n.size()-1);
		if(list_n.size()==0) ans -= 2*list_p.get(list_p.size()-1);
		System.out.println(ans);
		for(int i=0;i<n-1;i++) {
			if(list_n.size()==0 && n!=2) {
				long a = list_p.get(0);
				long b = list_p.get(list_p.size()-1);
				list_n.add(-a+b);
				list_p.remove(0);
				list_p.remove(list_p.size()-1);
				System.out.println(b + " " + a);
			}else if(list_n.size()==0 && n==2){
				long a = list_p.get(0);
				long b = list_p.get(list_p.size()-1);
				list_p.remove(0);
				list_p.remove(list_p.size()-1);
				list_p.add(a-b);
				System.out.println(a + " " + b);
			}else if(list_p.size()==0) {
				long a = list_n.get(0);
				long b = list_n.get(list_n.size()-1);
				list_p.add(-a+b);
				list_n.remove(0);
				list_n.remove(list_n.size()-1);
				System.out.println(b + " " + a);
			}else {
				long a = list_p.get(0);
				long b = list_n.get(0);
				if(list_p.size()<=list_n.size()) {
					list_p.add(a-b);
					System.out.println(a + " " + b);
				}else {
					list_n.add(-a+b);
					System.out.println(b + " " + a);
				}
				list_p.remove(0);
				list_n.remove(0);
			}
		}
		
		in.close();

	}

}