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>
#define rep(i,n) for(int i = 0; i < n; i++)
#define pb push_back
using namespace std;
typedef long long ll;

int main(){
  int n;
  cin >> n;
  vector<int> a(n);
  rep(i, n) cin >> a[i];
  vector<pair<int, int>> res;
  
  sort(a.begin(), a.end());
  int smallest = a[0];
  int biggest = a[a.size() - 1];
  for (int i = 1; i < n - 1; i++){
    if (a[i] < 0){
      res.pb({biggest, a[i]});
      biggest -= a[i];
    }else{
      res.pb({smallest, a[i]});
      smallest -= a[i];
    }
  }
  cout << biggest - smallest << endl;
  rep(i, (int)res.size()) cout << res[i].first << ' ' << res[i].second << endl;
  cout << biggest << ' ' << smallest << endl;
}