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;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CCYES cout<<"Yay!"<<endl
#define CCNO cout<<":("<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

int main(){
  int n;
  cin >> n;
  vector<int> a(n);
  bool hu=false,sei=false;
  rep(i,n){
    cin >> a[i];
    if(a[i]<=0) hu=true;
    if(a[i]>=0) sei=true;
  }
  sort(all(a));
  if(hu){
    if(sei){
      int m=0;
      rep(i,n) m+=abs(a[i]);
      cout << m << endl;
      int sum = a[0];
      int aa = a[n-1];
      repp(i,1,n-1){
        if(a[i]>0){
          cout << sum SP a[i] << endl;
          sum-=a[i];
        }else{
          cout << aa SP a[i] << endl;
          aa-=a[i];
        }
      }
      cout << aa SP sum << endl;
    }else{
      int m=a[n-1];
      rep(i,n){
        m+=abs(a[i]);
      }
      cout << m << endl;
      int sum = a[n-1];
      rep(i,n-1){
        cout << sum SP a[i] << endl;
        sum-=a[i];
      }
    }
  }else{
    int m=-a[0];
    repp(i,1,n){
      m+=abs(a[i]);
    }
    cout << m << endl;
    int sum=a[0];
    repp(i,2,n){
      cout << sum SP a[i] << endl;
      sum-=a[i];
    }
    cout << a[1] SP sum << endl;
  }
  return 0;
}