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 <bits/stdc++.h>

using namespace std;

vector<int>pos,neg;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n,i,x,tot=0;
    cin>>n;
    for(i=0;i<n;i++){
        cin>>x;
        if(x>=0)pos.push_back(x);
        else neg.push_back(abs(x));
    }
    if(n==1){cout<<x<<'\n';return 0;}
    sort(pos.begin(),pos.end());
    sort(neg.begin(),neg.end());
    if(neg.empty()){
        for(i=2;i<pos.size();i++)tot+=pos[i];
        tot += (pos[1]-pos[0]);
        cout<<tot<<'\n';
        if(n==2)cout<<pos[1]<<' '<<pos[0]<<'\n';
        else{
            cout<<pos[0]<<' '<<pos[1]<<'\n';
            int cur = pos[0] - pos[1];
            for(i=2;i<(int)pos.size()-1;i++){
                cout<<cur<<' '<<pos[i]<<'\n';
                cur-=pos[i];
            }
            cout<<pos[pos.size()-1]<<' '<<cur<<'\n';
        }
    }
    else if(pos.empty()){
        for(i=2;i<neg.size();i++)tot+=neg[i];
        tot += (neg[1]-neg[0]);
        cout<<tot<<'\n';
        cout<<neg[0]*(-1)<<' '<<neg[1]*(-1)<<'\n';
        int cur = neg[1] - neg[0];
        for(i=2;i<neg.size();i++){
            cout<<cur<<' '<<neg[i]*(-1)<<'\n';
            cur += neg[i];
        }
    }
    else{
        for(i=0;i<pos.size();i++)tot+=pos[i];
        for(i=0;i<neg.size();i++)tot+=neg[i];
        cout<<tot<<'\n';
        int cur = neg[0]*(-1);
        for(i=1;i<pos.size();i++){
            cout<<cur<<' '<<pos[i]<<'\n';
            cur-=pos[i];
        }
        cout<<pos[0]<<' '<<cur<<'\n';
        cur=pos[0]-cur;
        for(i=1;i<neg.size();i++){
            cout<<cur<<' '<<neg[i]*(-1)<<'\n';
            cur+=neg[i];
        }
    }
    return 0;
}