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);
  for(int i=0;i<n;i++){
    cin>>a[i];
  }
  sort(a.begin(),a.end());
  int index=distance(a.begin(),upper_bound(a.begin(),a.end(),-1));
  if(index==0){
    index=1;
  }else if(index==n){
    index=n-1;
  }
  int temp=a[0];
  int sum1=a[0],sum2=a[n-1];
  for(int i=index;i<n-1;i++){
    sum1-=a[i];
  }
  a[0]=sum1;
  for(int i=0;i<index;i++){
    sum2-=a[i];
  }
  cout<<sum2<<endl;
  a[0]=temp;
  sum1=a[0],sum2=a[n-1];
  for(int i=index;i<n-1;i++){
    cout<<sum1<<' '<<a[i]<<"\n";
    sum1-=a[i];
  }
  a[0]=sum1;
  for(int i=0;i<index;i++){
    cout<<sum2<<' '<<a[i]<<"\n";
    sum2-=a[i];
  }
  return 0;
}