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<iostream>
#include<algorithm>
#include<vector>


using namespace std;

#define ll long long
#define debug(n) cerr << #n << ':' << n << endl;

int n;
vector<int> vec;

// bool ok(int x){

//   int dif = vec[0];
//   for(int i = 1; i < n - 1; ++i)
//     dif -= vec[i];
//   int last = vec[n - 1] - dif;
//   // debug(x);
//   // debug(last);
//   return last >= x;
// }

// int maxBinarySearch(int  s, int  e, bool (*fp)(int )){
//   int  mid = (s + e) / 2;
//   if(e - s <= 1) return mid;
//   if(fp(mid))
//     return maxBinarySearch(mid, e, fp);
//   else
//     return maxBinarySearch(s, mid, fp);
// }


int main(){
  cin >> n;
  vec.resize(n);
  for(auto&&e : vec)cin >> e;
  sort(vec.begin(),vec.end());
  string out;
  int dif = vec[0];
  for(int i = 1; i < n - 1; ++i){
    out += to_string(dif);
    out += ' ';
    out += to_string(vec[i]);
    out += '\n';
    dif -= vec[i];
  }
  out += to_string(vec[n - 1]);
  out += ' ';
  out += to_string(dif);  
  dif = vec[n - 1] - dif;
  cout << dif << endl << out << endl;
  return 0;
}