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>
#include<iomanip>
#include<numeric>

using namespace std;
using ll = long long;
using ull = unsigned long long;
constexpr int mo = 1e9+7;
constexpr int  mod = mo;
constexpr int inf = 1<<30;
constexpr ll infl = 1ll<<60;
int main(){
    int n;
    cin>>n;
    vector<int> A(n);
    for(auto &a : A) cin>>a;
    sort(A.begin(),A.end());
    vector<pair<ll,ll>> ans;
    if(A[n-1] < 0){
        for(int i=n-2;i>=0;i--){
            ans.emplace_back(A[n-1],A[i]);
            A[n-1] -= A[i];
        }
        cout << A[n-1] << endl;
        for(auto an : ans){
            cout << an.first << " " << an.second << endl;
        }
    }else if(A[0] >= 0){
        for(int i=1;i<n-1;++i){
            ans.emplace_back(A[0],A[i]);
            A[0] -= A[i];
        }
        ans.emplace_back(A[1],A[0]);
        cout << A[1] - A[0] << endl;
        for(auto an : ans){
            cout << an.first << " " << an.second << endl;
        }
     }else {
        int count = lower_bound(A.begin(),A.end(),0) - A.begin();
        for(int i=count+1;i<n;++i){
            ans.emplace_back(A[count-1],A[i]);
            A[count-1] -= A[i];
        }
        for(int i=0;i<count-1;++i){
            ans.emplace_back(A[count],A[i]);
            A[count] -= A[i];
        }
        ans.emplace_back(A[count],A[count-1]);
        cout << A[count] - A[count-1] << endl;
        for(auto a : ans){
            cout << a.first << " " << a.second << endl;
        }
    }
    return 0;
}