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<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector> 
#include<limits>
#include<numeric>
#include<type_traits>
#include<math.h>
#include<fstream>
#include<set>
using namespace std;
typedef long long ll;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i = (int)(a); i < (int)(b); i++)
#define out(str) cout << str << endl
#define ALL(a) (a).begin(),(a).end()
#define INF (1<<29)

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

ll N,x,y,ans,now;

int main(){
    // ifstream in("input.txt");
    // cin.rdbuf(in.rdbuf());
    cin>>N;
    ll A[N];
    rep(i,N)cin>>A[i];
    sort(A,A+N);
    rep(i,N){
        if(i<N/2)ans-=A[i];
        else ans+=A[i];
    }
    out(ans);
    // out(A[0]);
    if(N%2){
        rep(i,N-1){
            if(i==N-2)cout<<A[N-1]<<" "<<now<<endl;
            else{
                if(i%2){
                    cout<<A[(i+1)/2]<<" "<<now<<endl;
                    now=now-A[(i+1)/2];
                }
                else{
                    if(i==0){
                        cout<<A[0]<<" "<<A[N/2]<<endl;
                        now=A[0]-A[N/2];
                    }
                    else{
                        cout<<A[N/2+i/2]<<" "<<now<<endl;
                        now=now-A[N/2+i/2];
                    }
                }
            }
        }
    }
    else{
        rep(i,N-1){
            if(i==N-2)cout<<A[N-1]<<" "<<-now<<endl;
            else{
                if(i%2){
                    cout<<A[(i+1)/2]<<" "<<now<<endl;
                    now=now-A[(i+1)/2];
                }
                else{
                    if(i==0){
                        cout<<A[N/2]<<" "<<A[0]<<endl;
                        now=A[N/2]-A[0];
                    }
                    else{
                        cout<<A[N/2+i/2]<<" "<<now<<endl;
                        now=now-A[N/2+i/2];
                    }
                }
            }
        }
    }
    
    

    
    
}