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;

int main()
{
  int n;
  cin >> n;
  vector<int> a(n);
  int i, j;
  for(i = 0; i < n; i++)
  {
    cin >> a.at(i);
  }
  sort(a.begin(), a.end());
  int p, q;
  p = 0;
  q = 0;
  for(i = 0; i < n; i++)
  {
    if(a.at(i) >= 0)p++;
    else q++;
  }
  if(p == 0)
  {
    p = 1;
    q = n-1;
  }
  if(q == 0)
  {
    p = n-1;
    q = 1;
  }
  int ans = 0;
  for(i = 0; i < n; i++)
  {
    if(i<q) ans -= a.at(i);
    else ans += a.at(i);
  }
  cout << ans << endl;
  for(i = q; i < n-1; i++)
  {
    cout << a.at(0) << ' ' << a.at(i) << endl;
    a.at(0) -= a.at(i);
  }
  for(i = 0; i < q; i++)
  {
    cout << a.at(n-1) << ' ' << a.at(i) << endl;
    a.at(n-1) -= a.at(i);
  }
}