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(){
	ios_base::sync_with_stdio(false); 
  cin.tie(NULL); cout.tie(NULL); 

  int n; cin >> n;
  vector < int > a(n);

  for(int &x : a) cin >> x;
  sort(a.begin(), a.end());

  long long ans = a.back() - a.front();
  for(int i = 1; i < a.size() - 1; ++i) {
    if(a[i] < 0) ans -= a[i];
    else ans += a[i];
  }

  cout << ans << endl;

  for(int i = 1; i < a.size() - 1; ++i) {
    if(a[i] < 0) {
      cout << a.back() << " " << a[i] << endl;
      a.back() -= a[i];
    }
    else {
      cout << a.front() << " " << a[i] << endl;
      a.front() -= a[i];
    }
  }

  cout << a.back() << " " << a.front() << endl;

  return 0;
}