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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int A[100001];

int main() {
  int N,M, max, min, sumM, sumP;
  int i, j, k;
  scanf ("%d", &N);
  
  for (i=0;i<N;i++) {
      scanf("%d", &A[i]);
  }
  
  max = -10000;
  min = 10000;
  for (i=0;i<N;i++) {
    if (max<A[i]) {max=A[i]; j=i;}
    if (min>A[i]) {min=A[i]; k=i;}
  }
  if (j==k) k=1;
  
  sumP=0;
  sumM=0;
  for (i=0;i<N;i++) {
    if (i!=j && i!=k) {
      if (A[i]>0) sumP+=A[i];
      if (A[i]<0) sumM+=A[i];
    }
  }
  M = sumP-sumM+max-min;
  printf ("%d\n", M);
  
  for (i=0;i<N;i++) {
    if (i!=j && i!=k) {
      if (A[i]>0) {printf("%d %d\n", min, A[i]); min=min-A[i];}
      else        {printf("%d %d\n", max, A[i]); max=max-A[i];}
    }
  }
  printf("%d %d\n", max, min);
  
  return 0;
}