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>
typedef long long LL;
using namespace std;

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());
  vector<int> x, y;
  int maxv = a.back(), minv = a.front();
  for (int i = 1; i < n - 1; i ++) {
    if (a[i] >= 0) {
      x.push_back(minv);
      y.push_back(a[i]);
      minv -= a[i];
    } else {
      x.push_back(maxv);
      y.push_back(a[i]);
      maxv -= a[i];
    }
  }
  x.push_back(maxv);
  y.push_back(minv);

  cout << maxv - minv << endl;
  for (int i = 0; i < x.size(); i ++) {
    cout << x[i] << " " << y[i] << endl;
  }
}