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> X(N,0);
  vector<pair<int,int>> P;
  for(int i=0;i<N;i++){
    cin>>X[i];
  }
  sort(X.begin(),X.end());
  
  for(int i=1;i<N-1;i++){
    if(X[i]>0){
      P.emplace_back(X[0],X[i]);
      X[0]-=X[i];
    }
    else{
      P.emplace_back(X[N-1],X[i]);
      X[N-1]-=X[i];
    }
  }
  
  /*cout<<X[0]-X[N-1]<<endl;
  
  for(int i=0;i<N-2;i++){
    cout<<setprecision(20)<<P[i].first<<" "<<P[i].second<<endl;
  }*/
  
    P.emplace_back(X[N-1],X[0]);
    cout << X[N-1] - X[0] << endl;
    for(auto &e : P){
        cout << e.first << " " << e.second << endl;
    }
  
  return 0;
}