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;
typedef long long ll;

int main(){
  ll N;
  cin >> N;
  vector<ll> A(N),q,w;
  for(int i=0;i<N;i++)cin >> A[i];
  sort(A.begin(),A.end());
  for(int i=1;i<N-1;i++){
    if(A[i]>=0){
      q.push_back(A[0]); w.push_back(A[i]);
      A[0]-=A[i];
    }
    else{
      q.push_back(A[N-1]); w.push_back(A[i]);
      A[N-1]-=A[i];
    }
  }
  q.push_back(A[N-1]); w.push_back(A[0]);
  cout << A[N-1]-A[0] << endl;
  int s=q.size();
  for(int i=0;i<s;i++)cout << q[i] << " " << w[i] << endl;
}