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>

using namespace std;

int N;
long long arr[100005];
vector<pair<long long, long long>> ans;

int main(){
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> N;
    long long mini = INT_MAX, maxi = INT_MIN, idxmini = 0, idxmaxi = 0;
    for(int i = 1; i<=N; i++){
        cin >> arr[i];
        if(arr[i] < mini){
            mini = arr[i];
            idxmini = i;
        }
        if(arr[i] >= maxi){
            maxi = arr[i];
            idxmaxi = i;
        }
    }
    if(maxi < 0){
        for(int i = 1; i<=N; i++){
            if(i != idxmaxi){
                ans.push_back({maxi, arr[i]});
                maxi -= arr[i];
            }
        }
    }
    else if(mini > 0){
        for(int i= 1; i<=N; i++){
            if(i != idxmini && i != idxmaxi){
                ans.push_back({mini, arr[i]});
                mini -= arr[i];
            }
        }
        ans.push_back({maxi, mini});
        maxi -= mini;
    }
    else{
        for(int i = 1; i<=N; i++){
            if(i != idxmaxi && arr[i] >= 0){
                ans.push_back({mini, arr[i]});
                mini -= arr[i];
            }
            if(i != idxmini && arr[i] < 0){
                ans.push_back({maxi, arr[i]});
                maxi -= arr[i];
            }
        }
        ans.push_back({maxi, mini});
        maxi -= mini;
    }
    cout << maxi << "\n";
    for(auto p : ans){
        cout << p.first << " " << p.second << "\n";
    }
}