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 <vector>
#include <array>
#include <set>
#include <cmath>
#include <string>
#include <queue>
#include <list>
#include <unordered_map>
#include <algorithm>
#include <functional>
#include <iomanip> // std::std::setprecision(X)
#include <limits>  //std::numeric_limits<int>::max(), min()

int main (void) {

  int N; std::cin >> N;
  std::vector<int> A(N);
  for(int i=0; i<N; i++) std::cin >> A[i];

  int st = 0, end = N-1;
  std::sort(A.begin(), A.end(), [] (int &x, int &y) { return x>y; });

  long long ans = 0;
  for(int i=1; i<N-1; i++) ans += std::abs(A[i]);
  ans += (A[st] - A[end]);
  std::cout << ans << std::endl;


  while(st != end) {
    int cminus = 0;
    for(int i=st; i<=end; i++) if ( A[i] < 0 ) cminus++;
    if ( cminus > 0 ) cminus--;

    if ( end - st == 1 ) {
      std::cout << A[st] << " " << A[end] << std::endl;
      A[st] -= A[end];
      end--;
    }
    else if ( cminus % 2 == 0 ) {
      std::cout << A[end] << " " << A[st] << std::endl;
      A[end] -= A[st];
      st++;
    }
    else {
      std::cout << A[st] << " " << A[end] << std::endl;
      A[st] -= A[end];
      end--;
    }
  }
  return 0;
}