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;
using ll = long long;
typedef pair<int, int> P;
ll Mod = 1000000007;
int main() {
  int N;
  cin >> N;
  int A[N];
  vector<int> plus;
  vector<int> minus;
  int zero = 0;
  ll sum = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    sum += abs(A[i]);
    if (A[i] == 0) {
      zero++;
    }
    if (A[i] > 0) {
      plus.push_back(A[i]);
    } else if (A[i] < 0) {
      minus.push_back(A[i]);
    }
  }

  
  if (plus.size() == N || minus.size() == N) {
    if (plus.size() == N) {
      sort(plus.begin(),plus.end());
      cout << sum - (plus[0] * 2) << endl;
      /*
      ll left = plus[plus.size() - 1];
      ll right = plus[0];
      for (int i = 0; i < plus.size() - 2; i++) {
        cout << right << " " << plus[i+1] << endl;
        right = right - plus[i+1];
      }
      cout << left << " " << right << endl;
      */
    } else {
      sort(minus.begin(),minus.end());
      reverse(minus.begin(),minus.end());
      cout << sum + (minus[0] * 2) << endl;
      /*
      ll left = minus[0];
      ll right = minus[minus.size() - 1];
      for (int i = 0; i < minus.size() - 2; i++) {
        cout << left << " " << minus[i+1] << endl;
        left = left - minus[i+1];
      }
      cout << left << " " << right << endl;
      */
    }
  } else {
    if (minus.size() == 0) {
      for (int i = 0; i < zero; i++) {
        minus.push_back(0);
      }
    } else {
      for (int i = 0; i < zero; i++) {
        plus.push_back(0);
      }
    }
    cout << sum << endl;
    /*
    ll res = plus[0];
    if (minus.size() > 1) {
      res = plus[0] - minus[1];
    }
    for (int i = 0; i < minus.size() - 1; i++) {
      if (i == 0) {
        cout << plus[0] << " " << minus[1] << endl;
      } else {
        cout << res << " " << minus[i+1] << endl;
        res = res - minus[i+1];
      }
    }
    ll res2 = minus[0];
    if (plus.size() > 1) {
      res2 = minus[0] - plus[1];
    }
    for (int i = 0; i < plus.size() - 1; i++) {
      if (i == 0) {
        cout << minus[0] << " " << plus[1] << endl;
      } else {
        cout << plus[i+1] << " " << res2 << endl;
        res2 = res2 - plus[i+1];
      }
    }
    cout << res << " " << res2 << endl;
    */
  }
  return 0;
}