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<algorithm>
#include<climits>
#include<cmath>
#include<cstdlib>
#include<iomanip>
#include<iostream>
#include<numeric>
#include<string>
#include<utility>
#include<vector>
using namespace std;

//--------------------

int main() {
	long res = 0;
	long resx[100009] = {}, resy[100009] = {}; long rin = 1;
	long N; cin >> N;
	long A[100009]; for (int i = 1; i <= N; i++) cin >> A[i];
	long abss[100009] = {};

	sort(A + 1, A + N + 1);

	if (N % 2 == 0) { //とりあえずNは偶数。
		abss[1] = A[N] - A[1]; resx[rin] = A[N]; resy[rin] = A[1]; rin++;//ここだけ特殊。引く順番が逆。
		for (int i = 2; i <= N / 2; i++) { abss[i] = A[i] - A[N + 1 - i]; resx[rin] = A[i]; resy[rin] = A[N + 1 - i]; rin++; }
		for (int i = 2; i <= N / 2; i++) { resx[rin] = abss[1]; resy[rin] = abss[i]; abss[1] = abss[1] - abss[i]; rin++; }//ここ書き換えに注意。
		res = abss[1];
        
	}
	if (N % 2 == 1) { //つぎNは奇数。
		for (int i = 1; i <= N / 2; i++) { abss[i] = A[i] - A[N + 1 - i]; resx[rin] = A[i]; resy[rin] = A[N + 1 - i]; rin++; }
		abss[0] = A[N / 2 + 1];
		for (int i = 1; i <= N / 2; i++) { resx[rin] = abss[0]; resy[rin] = abss[i]; abss[0] = abss[0] - abss[i]; rin++; }//ここ書き換えに注意。
		res = abss[0];
      cout << "Hi!!!" << endl;
	}


	cout << res << endl;
	for (int i = 1; i < rin; i++) cout << resx[i] << " " << resy[i] << endl;
}