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: Wrong Answer)

#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

int main() {
  int N;
  cin >> N;
  vector<ll> A(N);
  int minus = 0;
  ll sum = 0;
  ll mn = 1e9;
  ll mx = -1e9;
  REP(i, N) {
    cin >> A[i];
    if (A[i] < 0) minus++;
    sum += A[i];
    if (A[i] < mn) mn = A[i];
    if (mx < A[i]) mx = A[i];
  }
  ll ans1 = (sum - mn) - mn;
  ll ans2 = mx - (sum - mx);
  if (ans1 > ans2) {
    cout << ans1 << endl;
    sort(ALL(A));
    cout << A[0] << " " << A[1] << endl;
    ll tmp = A[0] - A[1];
    for (int i = 2; i < N - 1; ++i) {
      cout << tmp << " " << A[i] << endl;
      tmp -= A[i];
    }
    cout << mx << " " << tmp << endl;
  } else {
    cout << ans2 << endl;
    sort(ALL(A));
    reverse(ALL(A));
    cout << A[0] << " " << A[1] << endl;
    ll tmp = A[0] - A[1];
    for (int i = 2; i < N - 1; ++i) {
      cout << tmp << " " << A[i] << endl;
      tmp -= A[i];
    }
    cout << mx << " " << tmp << endl;
  }
  return 0;
}