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 <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define MOD 1000000007
typedef long long ll;
using namespace std;

int main(){
  int N;
  cin>>N;
  vector<ll> A(N);
  for(int i=0;i<N;i++) cin>>A[i];

  sort(A.begin(),A.end());

  vector<pair<ll,ll>> ans;
  int now=0;
  if(A[0]>0){
    ans.push_back(make_pair(A[0],A[N-1]));
    A[0]=A[0]-A[N-1];
    A.pop_back();
  }else if(A[N-1]<0){
    ans.push_back(make_pair(A[N-1],A[0]));
    now=1;
    A[N-1]=A[N-1]-A[0];
  }

  ll x=A.back();
  ll y=A[now];
  for(int i=now+1;i<A.size()-1;i++){
    if(A[i]<=0){
      ans.push_back(make_pair(x,A[i]));
      x=x-A[i];
    }else{
      ans.push_back(make_pair(y,A[i]));
      y=y-A[i];
    }
  }

  ans.push_back(make_pair(x,y));
  ll m=x-y;
  cout<<m<<endl;
  for(int i=0;i<ans.size();i++) cout<<ans[i].first<<" "<<ans[i].second<<endl;

  

  return 0;
}