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>
#include <vector>
#include <string>
using namespace std;

int main(){
  int N;
  cin >> N;
  vector<int> A(N);
  int sum;
  int max = N - 1;
  int min = 0;
  int ans = 0;

  for(int i = 0;i < N;i++){
    cin >> A[i];
  }

  sort(A.begin(),A.end());

  if(N % 2 == 0){
    for(int i = 0;i < N / 2;i++){
      ans -= A[i];  
    }
    for(int i = N / 2;i < N;i++){
      ans += A[i];
    }
  }else{
    for(int i = 0;i < N / 2;i++){
      ans -= A[i];  
      if(i == N / 2 && A[N / 2] < 0)ans += A[i] * 2;
    }
    for(int i = N / 2;i < N;i++){
      ans += A[i];
    } 
  }

  cout << ans << endl;

  for(int i = 0;i < N - 1;i++){
    if(i % 2 == 0){
      if(min == 0){
        cout << A[max] << " " << A[min] << endl;
        sum = A[max] - A[min];
        max--;
        min++;
      }else{
        cout << A[max] << " " << sum << endl;
        sum = A[max] - sum;
        max--;
      }
    }else{
      if(min == 0){
        cout << A[min] << " "<< A[max] << endl;
        sum = A[min] - A[max];
        max--;
        min++;
      }else{
        cout << A[min] << " "<< sum << endl;
        sum = A[min] - sum;
        min++;
      }
    }
  }

}