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>
int main(){
  int N;
  int Ap[100000];
  int Am[100000];
  int A;
  int i,j=0,k=0;
  scanf("%d",&N);
  for(int i=0;i<N;i++){
    scanf("%d",&A);
    if (A >= 0){
      Ap[j] = A;
      j++;
    }else if(A < 0){
      Am[k] = A;
      k++;
    }
  }
  int max = -20000;
  int min = 20000;
  int sum = 0;
  int tmp;
  int a,b;
  if (j == N){
    j = 0;
    for (j=0;j<N;j++){
      sum += Ap[j];
      if (Ap[j] < min){
        min = Ap[j];
        i = j;
      }
    }
    printf("%d\n",sum - 2*Ap[i]);
    j = 0;
    tmp = min;
    if (i == 0){
      for(j=2;j<N;j++){
        printf("%d %d\n",tmp,Ap[j]);
        tmp = tmp - Ap[j];
      }
      printf("%d %d\n",Ap[1],tmp);
    }else{
      for(j=1;j<N;j++){
        if(j != i){
          printf("%d %d\n",tmp,Ap[j]);
          tmp = tmp - Ap[j];
        }
      }
      printf("%d %d\n",Ap[0],tmp);
    }
  }else if(k == N){
    k = 0;
    for (k=0;k<N;k++){
      sum -= Am[k];
      if (Am[k] > max){
        max = Am[k];
        i = k;
      }
    }
    printf("%d\n",sum + 2*Am[i]);
    k = 0;
    tmp = max;
    for(k=0;k<N;k++){
      if(k != i){
        printf("%d %d\n",tmp,Am[k]);
        tmp = tmp - Am[k];
      } 
    }
  }else{
    for(a=0;a<j;a++){
      sum += Ap[a];
    }
    for(b=0;b<k;b++){
      sum -= Am[b];
    }
    printf("%d\n",sum);
    a = 1;
    b = 1;
    tmp = Am[0];
    for(a=1;a<j;a++){
      printf("%d %d\n",tmp,Ap[a]);
      tmp = tmp - Ap[a];
    }
    printf("%d %d\n",Ap[0],tmp);
    tmp = Ap[0] - tmp;
    for(b=1;b<k;b++){
      printf("%d %d\n",tmp,Am[b]);
    }
  }
  return 0;
}