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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
constexpr ll mod=1e9+7;

vector<pair<int,int>> output;
void process(int x,int y,vector<int>& vec_x,vector<int>& vec_y,vector<int>& tgt) {
  output.push_back({x,y});
  auto itr_x=lower_bound(vec_x.begin(), vec_x.end(),x);
  vec_x.erase(itr_x);
  auto itr_y=lower_bound(vec_y.begin(), vec_y.end(),y);
  vec_y.erase(itr_y);
  tgt.push_back(x-y);
}

int main()
{
  int n;
  cin>>n;
  vector<int> plus,minus;
  for (int i = 0; i < n; ++i)
  {
    int a;
    cin>>a;
    if (a>0) plus.push_back(a);
    else minus.push_back(a);
  }
  sort(plus.begin(), plus.end());
  sort(minus.begin(), minus.end(),[](auto lhs,auto rhs){
    return abs(lhs) < abs(rhs);
  });
  if (minus.empty()) {
    process(plus.front(),plus.back(),plus,plus,minus);
  } else if (plus.empty()) {
    process(minus.front(),minus.back(),minus,minus,plus);
  }

  while (plus.size()!=1) {
    process(minus.front(),plus.front(),minus,plus,minus);
  }
  while (!minus.empty()) {
    process(plus.front(),minus.front(),plus,minus,plus);
  }
  cout<<plus.front()<<endl;
  for (int i = 0; i < output.size(); ++i)
  {
    cout<<output[i].first<<" "<<output[i].second<<endl;
  }
}