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;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
typedef pair<int, int> P;

int main()
{
  int n;
  cin >> n;
  deque<int> deq(n);
  rep(i, n) cin >> deq[i];
  sort(deq.begin(), deq.end(), greater<int>());
  
  int cur = deq[0];
  deq.pop_front();
  vector<P> sousa(0);
  while(deq.size())
  {
    int a = deq.front();
    if(a < 0)
    {
      sousa.emplace_back(P(cur, a));
      cur -= a;
      deq.pop_front();
    }
    else
    {
      if(cur > 0)
      {
        sousa.emplace_back(P(cur, a));
        a = deq.back();
        deq.pop_back();
        cur = a - cur;
      }
      else
      {
        sousa.emplace_back(P(a, cur));
        cur = a - cur;
        deq.pop_front();
      }
    }
    //cout << cur << endl;
  }
  
  cout << cur << endl;
  for(auto x : sousa) cout << x.first << " " << x.second << "\n";
  
  return 0;
}