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 <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0); cin.sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  int m, sum;
  for (int i = 0; i < n; i++) {
    sum += a[i] > 0 ? a[i] : -a[i];
  }
  vector<pair<int, int>> ret;
  int head = a[0];
  int tail = a[n - 1];
  if (head < 0) {
    m = sum + 2 * head;
    for (int i = 1; i < n; i++) {
      ret.emplace_back(a[0], a[i]);
      a[0] -= a[i];
    }
  } else if (tail > 0) {
    m = sum - 2 * tail;
    for (int i = 1; i < n - 1; i++) {
      ret.emplace_back(a[n - 1], a[i]);
      a[n - 1] -= a[i];
    }
    ret.emplace_back(a[0], a[n - 1]);
  } else {
    m = sum;
    int k = n;
    for (int i = 0; i < n; i++) {
      if (a[i] < 0) k = i;
    }
    for (int i = 1; i < k; i++) {
      ret.emplace_back(a[k], a[i]);
      a[k] -= a[i];
    }
    for (int i = k; i < n; i++) {
      ret.emplace_back(a[0], a[i]);
      a[0] -= a[i];
    }
  }
  cout << m << endl;
  for (auto e : ret) {
    cout << e.first << " " << e.second << endl;
  }
  return 0;
}