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++ (GCC 9.2.1) (judge verdict: Accepted)

//ヘッダー
#include<bits/stdc++.h>
using namespace std;

//型定義
typedef long long ll;

//定数
const int INF=1e+9;
const int MOD=1e+9+7;

//REPマクロ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define REP2(i,a,b) for(ll i=a;i<(ll)(b);i++)
#define REPD2(i,a,b) for(ll i=a;i>(ll)(b);i--)

// 多次元 vector 生成
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

//vectorの扱い
#define ALL(x) (x).begin(),(x).end() //sortなどの引数省略
#define SIZE(x) ((ll)(x).size()) //size
#define MAX(x) *max_element(ALL(x)) //最大値
#define MIN(x) *min_element(ALL(x)) //最小値

int main(){
    ll N;
    cin>>N;
    vector<ll> A(N),A_minus,A_plus;
    REP(i,N){
        cin>>A[i];
        if(A[i]>=0) A_plus.push_back(A[i]);
        else A_minus.push_back(A[i]);

    }
    sort(ALL(A));
    sort(ALL(A_minus));
    sort(ALL(A_plus));
    
    ll sum=0;
    vector<pair<ll,ll>> ans(N-1);
    if(A_minus.size()==0){
        ll x,y;
        x=A[0];
        y=A[1];
        ans[0]=make_pair(x,y);
        REP2(i,1,N-1){
            x-=y;
            y=A[i+1];
            if(i==N-2){
                sum=y-x;
                ans[i]=make_pair(y,x);
                break;
            }
            ans[i]=make_pair(x,y);   
        }
    }else if(A_plus.size()==0){
        ll x,y;
        sort(ALL(A),greater<ll>());
        x=A[0];
        y=A[1];
        ans[0]=make_pair(x,y);
        REP2(i,1,N-1){
            x-=y;
            y=A[i+1];
            ans[i]=make_pair(x,y);   
        }
        sum=x-y;
    }else{
        ll x,y;
        //マイナス1個とプラス多数
        x=A_minus[0];
        REP(i,A_plus.size()-1){
            y=A_plus[i];
            ans[i]=make_pair(x,y);
            x-=y;
        }
        ll y_last=x;
        //プラス1個とマイナス多数
        x=A_plus[A_plus.size()-1];
        REP2(i,1,A_minus.size()){
            y=A_minus[i];
            ans[A_plus.size()-2+i]=make_pair(x,y);
            x=x-y;
        }
        ans[N-2]=make_pair(x,y_last);
        sum=x-y_last;

    }
    cout<<sum<<endl;
    REP(i,N-1){
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }

}