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>
typedef long long ll;
using namespace std;

int main() {
  int n;
  cin>>n;
  ll a[100005];
  for (int i=0; i<n; i++) {
    cin>>a[i];
  }
  sort(a,a+n);
  int x[100005],y[100005];
  int i=0;
  int tmp1,tmp2;
  if (a[n-2]>=0) {
    x[0]=a[0];
    y[0]=a[n-2];
    for (i=1; i<n-2; i++) {
      if (a[n-2-i]<0) break;
      x[i]=x[i-1]-y[i-1];
      y[i]=a[n-2-i];
    }
    tmp1=x[i-1]-y[i-1];
  }
  else tmp1=a[0];
  if (i<n-2) {
    x[i]=a[n-1];
    y[i]=a[n-2-i];
    i++;
    for (; i<n-2; i++) {
      x[i]=x[i-1]-y[i-1];
      y[i]=a[n-2-i];
    }
    tmp2=x[n-3]-y[n-3];
  }
  else tmp2=a[n-1];
  x[n-2]=tmp2;
  y[n-2]=tmp1;

  cout<<x[n-2]-y[n-2]<<endl;
  for (int i=0; i<n-1; i++) {
    cout<<x[i]<<' '<<y[i]<<endl;
  }
  return 0;
}

