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 <iostream>
#include <numeric>
#include <vector>

using namespace std;

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<int> x, y;
  for (int i : a) {
    if (i < 0) {
      x.emplace_back(i);
    } else {
      y.emplace_back(i);
    }
  }
  if (y.empty()) {
    auto b = a;
    for (int &i : b) {
      i = abs(i);
    }
    cout << accumulate(b.begin(), b.end(), 0) - 2 * b[0] << endl;
    int t = a.back();
    for (int i = 0; i < n - 1; ++i) {
      cout << t << " " << a[i] << endl;
      t -= a[i];
    }
  } else if (a[0] < 0) {
    auto b = a;
    for (int &i : b) {
      i = abs(i);
    }
    cout << accumulate(b.begin(), b.end(), 0) << endl;
    int t = x[0];
    for (int i = 0; i < int(y.size()) - 1; ++i) {
      cout << t << " " << y[i] << endl;
      t -= y[i];
    }
    cout << y.back() << " " << t << endl;
    t = y.back() - t;
    for (int i = 1; i < int(x.size()); ++i) {
      cout << t << " " << x[i] << endl;
      t -= x[i];
    }
  } else {
    cout << accumulate(a.begin(), a.end(), 0) - 2 * a[0] << endl;
    int t = y[0];
    for (int i = 1; i < int(y.size()) - 1; ++i) {
      cout << t << " " << y[i] << endl;
      t -= y[i];
    }
    cout << y.back() << " " << t << endl;
  }
}