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 P pair<int, int>

using namespace std;
const int INF=1e5;

int main(){
  long long n, a[100009];
  cin >> n;
  for(int i=0;i<n;i++)cin >> a[i];
  sort(a, a+n);
  long long ans=a[n-1]-a[0];
  for(int i=1;i<n-1;i++)ans+=abs(a[i]);
  cout << ans << endl;
  for(int i=1;i<n-1;i++){
    if(a[i]>=0){
      cout << a[0] << ' ' << a[i] << endl;
      a[0]-=a[i];
    }
    else{
      cout << a[n-1] << ' ' << a[i] << endl;
      a[n-1]-=a[i];
    }
  }
  cout << a[n-1] << ' ' << a[0] << endl;
}

