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: Accepted)

using namespace std;

#include <iostream>
#include <vector>
#include <set>
#include <string>
#include <algorithm>
#include <utility>
#include <functional>
#include <queue>
#include <cmath>
#include <numeric>
#include <map>
using namespace std;


int main(void){
	int N,M;

	cin >> N;

  multiset<long long int> A, plu, minu;
  for (int i = 0;i<N;i++)
  {
    long long int tmp;
    cin >> tmp;
    A.insert(tmp);
    if (tmp <= 0)
      minu.insert(tmp);
    else
      plu.insert(tmp);
  }
  
  auto sum = 0LL, neuv = 0LL;
  vector<pair<long long int,long long int>> r;
  
  while (plu.size() + minu.size() > 1) {
    if (plu.empty()){
      if (minu.size() == 2){
      	neuv = *minu.rbegin() - *(minu.rbegin()++);
     	r.emplace_back(*minu.rbegin(), *(minu.rbegin()++));
      }
      else{
      	neuv = *(minu.rbegin()++) - *minu.rbegin();
     	r.emplace_back(*(minu.rbegin()++), *minu.rbegin());
      }
      sum = neuv;
      minu.erase(--minu.end());
      minu.erase(--minu.end());
      if (neuv < 0)
        minu.insert(neuv);
      else
        plu.insert(neuv);
    }
    else if (minu.empty()) {
      if (plu.size() == 2){
        neuv = *(plu.begin()++) - *plu.begin();
      r.emplace_back(*(plu.begin()++), *plu.begin());
      }
      else{
        neuv = *plu.begin() - *(plu.begin()++);
      r.emplace_back(*plu.begin(), *(plu.begin()++));
      }
      sum = neuv;
      plu.erase(plu.begin());
      plu.erase(plu.begin());
      if (neuv <= 0)
        minu.insert(neuv);
      else
        plu.insert(neuv);
    }
    else if (plu.size() <= minu.size()) {
      neuv = *plu.begin() - *minu.rbegin();
      sum = neuv;
      r.emplace_back(*plu.begin(), *minu.rbegin());
      plu.erase(plu.begin());
      minu.erase(--minu.end());
      if (neuv < 0)
        minu.insert(neuv);
      else
        plu.insert(neuv);
      
    }
    else {
      neuv = *minu.rbegin() - *plu.begin();
      sum = neuv;
       r.emplace_back(*minu.rbegin(), *plu.begin());
      plu.erase(plu.begin());
      minu.erase(--minu.end());
      if (neuv <= 0)
        minu.insert(neuv);
      else
        plu.insert(neuv);
      
    }
  }
	cout << sum << endl;
  for (auto&& i : r)
    cout << i.first << " " << i.second << endl;

	return 0;
}