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 <bits/stdc++.h>
using namespace std;
using P=pair<int,int>;
int main(){
  int n; cin>>n;
  priority_queue<int> pq,pq2;
  for(int i=0;i<n;i++){
    int a; cin>>a;
    pq.push(a); a*=-1;
    pq2.push(a);
  }
  vector<P> ans(n-1);
  bool flag=false;
  if(n%2==0) flag=true;
  int i=0;
  while(i!=n-1){
    if(flag){
      int x=pq.top()+pq2.top();
      ans[i]=P(pq.top(),-1*pq2.top());
      pq.pop(); pq2.pop();
      pq.push(x); pq2.push(-x);
      flag=false;
    }else{
      int x=-pq2.top()-pq.top();
      ans[i]=P(-1*pq2.top(),pq.top());
      pq.pop(); pq2.pop();
      pq.push(x); pq2.push(-x);
      flag=true;
    }
    i++;
  }
  cout<<pq.top()<<endl;
  for(int i=0;i<n-1;i++){
    cout<<ans[i].first<<' '<<ans[i].second<<endl;
  }
}