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 <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 10;

int main()
{
  int N; cin >> N;

  vector<int> A(N);
  for (int i = 0; i < N; i++) cin >> A[i];
  sort(A.begin(), A.end());

  int ans = -A[0];
  for (int i = 1; i < N - 1; i++)
  {
    if (A[i] < 0) ans -= A[i];
    else ans += A[i];
  }
  ans += A[N - 1];
  cout << ans << endl;

  int now = A[0];
  for (int i = N - 2; i >= 1; i--)
  {
    if (A[i] < 0) break;
    cout << now << " " << A[i] << endl;
    now = now - A[i];
  }

  cout << A[N - 1] << " " << now << endl;
  now = A[N - 1] - now;

  for (int i = 1; i < N - 1; i++)
  {
    if (A[i] >= 0) break;
    cout << now << " " << A[i] << endl;
    now = now - A[i];
  }
}