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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define rep2(i, s, n) for (ll i = (s); i < (ll)(n); i++)
#define all(v) v.begin(), v.end()
#define sz(v) v.size()
#define INF 100000000000000 //10^14
template <typename T>
bool chmax(T &a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

int main()
{
    ll N;
    cin >> N;
    vector<ll> A(N);
    rep(i, N) cin >> A.at(i);
    sort(all(A));
    ll p = N;
    rep(i, N){
      if(A.at(i) > 0){
        p = i;
        break;
      }
    }
    ll M = 0;
    M += A.at(N-1);
    M -= A.at(0);
    rep(i, N-2){
      if(i+1 >= p){
        M += A.at(i+1);
      }
      else{
        M -= A.at(i+1);
      }
    }
    cout << M << endl;
    ll x = A.at(0);
    if(p == 0){
      p++;
    }
    rep2(j, p, N-1){
      cout << x << " " << A.at(j) << endl;
      x -= A.at(j);
    }
    ll y = A.at(N-1);
    rep2(i, 1, p){
      cout << y << " " << A.at(i) << endl;
      y -= A.at(i);
    }
    cout << y << " " << x << endl;
}