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++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
ID: skipian1
PROB: 
LANG: C++11
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF = 2000000000
#define sz(a) int((a).size())
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define X first
#define Y second

int main() {
  int n, cut;
  cin >> n;
  int arr[n];
  int sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += abs(arr[i]);
  }
  sort(arr, arr+n);
  if (arr[0] <= 0 && arr[n-1] >= 0) {
    cout << sum << "\n";
    cut = n-1;
    for (int i = 1; i < n-1; i++) {
      if (arr[i] >= 0) {
	cut = i;
	break;
      }
    }
    int cur = arr[0];
    for (int i = cut; i < n-1; i++){ 
      cout << cur << " " << arr[i] << "\n";
      cur -= arr[i];
    }
    cout << arr[n-1] << " " << cur << "\n";
    cur = arr[n-1] - cur;
    for (int i = 1; i < cut; i++) {
      cout << cur << " " << arr[i] << "\n";
      cur -= arr[i];
    }
  } else if (arr[n-1] < 0) {
    cout << sum+2*arr[n-1] << "\n";
    int cur = arr[n-1];
    for (int i = n-2; i >= 0; i--) {
      cout << cur << " "  << arr[i] << "\n";
      cur -= arr[i];
    }
  } else {
    cout << sum-2*arr[0] << "\n";
    int cur = arr[0];
    for (int i = 1; i < n-1; i++) {
      cout << cur << " "  << arr[i] << "\n";
      cur -= arr[i];
    }
    cout << arr[n-1] << " " << cur << "\n";
  }
}