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> an(n);
  for(int i=0; i<n; ++i) cin >> an[i];
  sort(an.begin(),an.end());
  int ans = 0;
  for(int i=0; i<n; ++i){
    if(i == 0 && an[i] > 0) ans -= an[0];
    else if(i == n-1 && an[i] < 0) ans += an[n-1]; 
    else ans += abs(an[i]);
  }
  cout << ans << endl;
  for(int i=n-2; an[i]>=0 && i>0; --i){
    printf("%d %d\n", an[0], an[i]);
    an[0] -= an[i];
  }
  for(int i=1; an[i]<0 && i<n-1 ; ++i){
    printf("%d %d\n", an[n-1], an[i]);
    an[n-1] -= an[i];
  }
  printf("%d %d\n", an[n-1], an[0]);
}