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: Wrong Answer)

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

#define N_MAX 100000

int desc(const void *a, const void *b){
  if(*(int *)a > *(int *)b){
    return -1;
  }else if(*(int *)a == *(int *)b){
    return 0;
  }else{
    return 1;
  }
}

int main(void){
  int i, j;

  int n;
  scanf("%d", &n);

  int a[N_MAX] = {}, plus = 0, minus = 0;
  long sum = 0;
  for(i = 0; i < n; i++){
    scanf("%d", &a[i]);
    if(a[i] > 0){
      plus++;
      sum += a[i];
    }else if(a[i] < 0){
      minus++;
      sum -= a[i];
    }
  }

  qsort(a, n, sizeof(int), desc);

  if(plus == 0) sum += a[0]*2;
  else if(minus == 0) sum -= a[n-1]*2;
  printf("%ld\n", sum);

  for(i = 0; i < plus-1; i++){
    printf("%d %d\n", a[n-1], a[i]);
    a[n-1] -= a[i];
  }
  for(i = plus; i< n; i++){
    printf("%d %d\n", a[plus-1], a[i]);
  }

  return 0;
}