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>
#include <vector>
#include <algorithm>
using namespace std;
#define rep(i, start, end) for (int i = start; i < end; i++)


int main(){
  int m;

  cin >> m;
  vector<int> a;
  int x;
  int b[m+1];
  rep(i, 0, m) {
    cin >> x;
    a.push_back(x);
  }
  sort(a.begin(), a.end());

  b[0] = a[0];
  rep(i, 0, m-1) {
    b[i + 1] = b[i] - a[m - 1 - i];
  }
  cout << abs(b[m-1]) << endl;
  cout << b[0] << " " << a[m - 1] << endl;
  rep(i, 1, m-1) {

    if (b[i] >= a[m - (i+1)]) {
      cout << b[i] << " " << a[m - (i+1)] << endl;
    } else {
      cout << a[m - (i+1)] << " " << b[i] << endl;
    }
  }

  return 0;
}