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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

int main()
{
	int n,i,j,k;	
	cin >> n;
	int a[100001];
	int b[200002];
	for (i = 0; i < n; i++)
	{
		cin >> a[i];		
	}

	for (i = 0; i < n; i++)
	{
		for (j = i+1; j < n; j++)
		{
			if(a[i]>a[j])
			{
				a[i] = a[i] + a[j];
				a[j] = a[i] - a[j];
				a[i] = a[i] - a[j];
			}
			
		}
		b[i] = a[i];
	}
	
	for (i = 0; i < n - 2; i++)
	{
		b[2 * i] = a[0];
		b[2 * i + 1] = a[n - i - 1];
		a[0] = a[0] - a[n - i - 1];
		
	}
	cout << a[1] - a[0]; cout << endl;
	
	for (i = 0; i < n - 2; i++)
	{
		cout << b[2*i] <<" "<< b[2*i+1] << endl;		
	}
	cout << a[1] <<" "<< a[0];
}