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.Scanner;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.Collections;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a =new int[n];
		boolean high=false;
		boolean low=false;
		boolean zero=false;
		long sum=0;
		int zz=0;
		int ll=0;
		ArrayList<Integer> llist=new ArrayList<>();
		ArrayList<Integer> hlist=new ArrayList<>();
		int zcount=0;
		for(int i=0; i<n; i++){
			a[i]=sc.nextInt();
			if(a[i]>0){
				sum+=a[i];
				high=true;
				hlist.add(a[i]);
			}else if(a[i]<0){
				sum+=-a[i];
				low=true;
				ll=a[i];
				llist.add(a[i]);
			}else{
				zero=true;
				zz=a[i];
				zcount++;
			}
		}
		if(zero || (high && low)){
			System.out.println(sum);
			if(zero && !high && n-zcount>1 && low){
				System.out.println(0+" "+llist.get(0));
				hlist.add(-llist.remove(0));
				zcount--;
			}else if(zero && !low && n-zcount>1 && high){
				System.out.println(0+" "+hlist.get(0));
				llist.add(-hlist.remove(0));
				zcount--;
			}
			while(zcount>0){
				zcount--;
				System.out.println(hlist.get(0)+" "+0);
			}
		}else{
			Arrays.sort(a);
			sum-=a[0]+a[1];
			sum+=Math.abs(a[0]-a[1]);
			System.out.println(sum);
			if(high){
				Collections.sort(hlist);
				llist.add(hlist.get(0)-hlist.get(1));
				System.out.println(hlist.remove(0)+" "+hlist.remove(1));
			}else{
				Collections.sort(llist);
				hlist.add(llist.get(llist.size()-1)-llist.get(llist.size()-2));
				System.out.println(llist.remove(llist.size()-1)+" "+llist.remove(llist.size()-2));
			}
		}
			int count=0;
			while(llist.size()>1 &&hlist.size()>0){
				hlist.add(hlist.get(0)-llist.get(count));
				System.out.println(hlist.remove(0)+" "+llist.remove(count++));
			}
			count=0;
			while(hlist.size()>1 && llist.size()>0){
				llist.add(llist.get(0)-hlist.get(count));
				System.out.println(llist.remove(0)+" "+hlist.remove(count++));
			}
		if(hlist.size()+llist.size()>1){
			System.out.println(hlist.get(0)+" "+llist.get(0));
		}
			
		
		
	}
}