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;
using P= pair<int,int>;

int main(){
  int n; cin>>n;
  int num[n];
  for(int i=0;i<n;i++){
    cin>>num[i];
  }
  sort(num,num+n);
  vector<P> ans;
  int max=num[n-1],min=num[0];
  for(int i=1;i<n-1;i++){
    if(num[i]<0){
      ans.push_back(P(max,num[i]));
      max-=num[i];
    }else{
      ans.push_back(P(min,num[i]));
      min-=num[i];
    }
  }
  ans.push_back(P(max,min));
  cout<<max-min<<endl;
  for(int i=0;i<n-1;i++){
    cout<<ans[i].first<<' '<<ans[i].second<<endl;
  }
}