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<iostream>
#include<algorithm>
#include<vector>
#include<utility>
#include<cmath>
using namespace std;
#define pb push_back
typedef long long ll;
#define rep(i, a, b) for(ll i=(a); i<(b); i++)

ll a[100000];
int main(){
  ll N,div=0,sum = 0, adjust=0;
  vector<pair<ll, ll> > res;
  vector<ll> later;
  cin >> N;
  rep(i,0,N){
    cin>> a[i];
    sum += abs(a[i]);
  }
  sort(a,a+N);
  if(N == 2){
    cout<< a[1]-a[0]<<endl;
    cout << a[1] <<" " <<a[0]<<endl;
    return 0;
  }

  
  if(a[0] < 0 && a[N-1] >= 0){ 
    rep(i,1,N-1){
      if(a[i] > 0){
	res.pb(make_pair(a[0],a[i]));
	a[0] -= a[i];
      }else later.pb(i);
    }
    later.pb(0);
    rep(i,0,later.size()){
      res.pb(make_pair(a[N-1],a[later[i]]));
      a[N-1] -= a[later[i]];
    }
  }

  
  if(a[0]<0 && a[N-1] < 0){
    sum = abs(sum);
    sum += a[N-1] * 2;
    rep(i,0,N-1){
      res.pb(make_pair(a[N-1],a[i]));
      a[N-1] -= a[i];
    }
  } 
  if(a[0]>=0 && a[N-1] >=0){
 
    sum -= (a[0] * 2);
    rep(i,1,N-1){
      res.pb(make_pair(a[0],a[i]));
      a[0] -= a[i];
    }
    res.pb(make_pair(a[N-1],a[0]));
    a[N-1] -=a[0];
  }

  cout<< sum << endl;
  rep(i,0,res.size()){
    cout << res[i].first << " " << res[i].second << endl;
  }
}