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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <cstring>
#include <map>
#include <set>
#include <queue>
#include <math.h>
#include <algorithm>

using namespace std;

int main()
{
  int N;
  cin >> N;
  vector<int> A(N,0);
  for (int i=0;i<N;i++) cin >> A[i];
  sort(A.begin(), A.end());
  int maxa = A[N-1];
  int mina = A[0];
  vector<pair<int, int>> Pans;
  for (int i=1;i<N-1;i++) {
    if (A[i] < 0) {
      Pans.push_back(make_pair(maxa, A[i]));
      maxa -= A[i];
      continue;
    }
    Pans.push_back(make_pair(mina, A[i]));
    mina -= A[i];
  }
  Pans.push_back(make_pair(maxa, mina));
  cout << maxa - mina << endl;
  for (auto& v : Pans)
  {
    printf("%d %d\n", v.first, v.second);
  }
  return 0;
}
