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 <iostream>
#include<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<deque>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>

#define rep(i, n) for(int i=0;i<n;i++)

typedef int long long ll;
using namespace std;
typedef pair<int, int> P;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const ll MOD = 1e9 + 7;
static const int MAX = 100;
static const int INF = (1 << 23);



int main() {
    ll n;
    cin >> n;
    vector<ll> v(n);
    vector<ll> m;
    vector<ll> s;
    bool f = true;
    bool t = true;


    rep(i, n) {
        cin >> v[i];
        if (v[i] > 0)f = false;
        if (v[i] < 0)t = false;
        if (v[i] >= 0)s.push_back(v[i]);
        if (v[i] < 0)m.push_back(v[i]);
    }
    //vector<ll> c;
    //sort(c.begin(),c.end());
    sort(v.begin(), v.end());
    sort(s.begin(), s.end());
    sort(m.begin(), m.end());
    vector<pair<ll, ll>> ans;


    if(n==2){
        cout<<v[1]-v[0]<<endl;
        cout<<v[1]<<" "<<v[0]<<endl;
        return 0;
    }


    if (f) {//負のみ


        ll las = m[m.size() - 1];
       // cout<<las<<endl;
        for (int i = 0; i < m.size()-1 ; i++) {
            ans.push_back({las, m[i]});
            las -= m[i];
        }

        cout << las << endl;

        rep(i, ans.size()) {
            cout << ans[i].first << " " << ans[i].second << endl;
        }



    }else if(t){//正のみ

            ll lass=s[0];
            for(int i=2;i<s.size();i++){
                ans.push_back({lass,s[i]});
                lass-=s[i];
            }
           // cout<<lass<<endl;


            ans.push_back({s[1],lass});
            cout<<s[1]-lass<<endl;
            rep(i,ans.size()){
                cout<<ans[i].first<<" "<<ans[i].second<<endl;
            }

        


     }else{

        ll las=m[m.size()-1];
        for(int i=1;i<s.size();i++){
            ans.push_back({las,s[i]});
            las-=s[i];
        }

        ll las2=s[0];
        for(int j=0;j<m.size()-1;j++){
            ans.push_back({las2,m[j]});
            las2-=m[j];
        }


        ans.push_back({las2,las});
        cout<<las2-las<<endl;
        rep(i,ans.size()){
            cout<<ans[i].first<<" "<<ans[i].second<<endl;
        }


    }










    return 0;
}