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)

#include <bits/stdc++.h>

#define mp make_pair
#define fst first
#define snd second
#define forn(i,n) for (int i = 0; i < int(n); i++)
#define forn1(i,n) for (int i = 1; i <= int(n); i++)
#define popcnt __builtin_popcount

using namespace std;

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
typedef pair<int,ll> pil;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;

template <typename T> T id(T b) {return b;};
template <class It> bool all(It f,It l){return std::all_of(f,l,id<bool>);}
template <class It> bool any(It f,It l){return std::any_of(f,l,id<bool>);}
template <typename T> void chmax(T &x, T y) {if (x < y) x = y;}
template <typename T> void chmin(T &x, T y) {if (x > y) x = y;}

const int MAX_N = 100000;
int n;
int a[MAX_N];

int main()
{
#ifdef FASTIO
  ios_base::sync_with_stdio(false); cin.tie(nullptr);
#endif

  cin >> n;
  forn(i,n) cin >> a[i];

  sort(a,a+n);

  int res = -a[0] + a[n-1];
  for (int i = 1; i < n-1; i++)
    res += abs(a[i]);

  cout << res << endl;

  int i = n-2;
  int prev = a[0];
  while (i > 0 && a[i] >= 0) {
    cout << prev << " " << a[i] << "\n";
    prev -= a[i];
    i--;
  }

  cout << a[n-1] << " " << prev << "\n";
  prev = a[n-1] - prev;

  while (i > 0) {
    cout << prev << " " << a[i] << "\n";
    prev -= a[i];
    i--;
  }

  return 0;
}