Problem:
Score : 500 points

Problem Statement

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}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int main()
{
  int n,a[10000];
  scanf("%d",&n);
  for(int i=0;i<n;i++)
  {
    scanf("%d",&a[i]);
  }
  for(int i=0;i<n;i++)
  {
    for(int j=i+1;j<n;j++)
    {
      if(a[i]>a[j])
      {
        int temp=a[j];
        a[j]=a[i];
        a[i]=temp;
      }
    }
  }
    printf("%d\n",a[n-1]-(a[0]-a[1]));
    printf("%d %d\n",a[0],a[1]);
    printf("%d %d",a[n-1],a[0]-a[1]);
  return 0;
  }