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 argc, char const *argv[]) {
  int N,arr[100000],i,j,tmp,ans=0,arr2[100000],flg=0;
  scanf("%d", &N);

  for (i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
  }

  for (i = 0; i < N; i++) {
    /* code */
  }

  for (i = 0; i < N; i++) {
    for (j = i+1; j < N; j++) {
      if (arr[i] > arr[j]) {
        tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
      }
    }
  }

  if (N%2 == 0) {
    for (i = 0; i < N/2; i++) {
      if (flg == 0) {
        arr2[i] = arr[N-1-i] - arr[i];
        flg = 1;
      } else {
        arr2[i] = arr[i] - arr[N-1-i];
      }
    }
    ans += arr2[0];
    for (j = 1; j < i; j++) {
      ans -= arr[j];
    }
  } else {
    for (i = 0; i < N/2; i++) {
      arr2[i] = arr[i] - arr[N-1-i];
    }
    arr2[i] = arr[N/2];
    for (j = 0; j < i; j++) {
      ans -= arr[j];
    }
    ans += arr[j];
  }

  printf("%d\n", ans);

  if (N%2 == 0) {
    for (i = 0; i < N/2; i++) {
      if (flg == 0) {
        printf("%d %d\n", arr[N-1-i], arr[i]);
        arr2[i] = arr[N-1-i] - arr[i];
        flg = 1;
      } else {
        printf("%d %d\n", arr[i], arr[N-1-i]);
        arr2[i] = arr[i] - arr[N-1-i];
      }
    }
    for (j = 0; j < i-1; j++) {
      printf("%d %d\n", arr2[j], arr2[j+1]);
      arr2[j+1] = arr2[j] - arr2[j+1];
    }
  } else {
    for (i = 0; i < N/2; i++) {
      if (flg == 0) {
        printf("%d %d\n", arr[N-1-i], arr[i]);
        arr2[i] = arr[N-1-i] - arr[i];
        flg = 1;
      } else {
        printf("%d %d\n", arr[i], arr[N-1-i]);
        arr2[i] = arr[i] - arr[N-1-i];
      }
      flg = 1;
    }
    arr2[i] = arr[N/2];
    for (j = 0; j < i-1; j++) {
      printf("%d %d\n", arr2[j], arr2[j+1]);
    }
    printf("%d %d\n", arr2[j+1], arr2[j]);
  }

  return 0;
}