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)

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

//#define int long long
typedef long long ll;
//const int INF = 2e9;
//const ll INF = 9e18;


signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int N;
    cin >> N;
    vector<int> A(N);
    for (int i=0;i<N;i++) cin >> A[i];
    sort(A.begin(),A.end());
    vector<int> plus;
    vector<int> minus;
    plus.push_back(A[N-1]);
    minus.push_back(A[0]);
    for (int i=1;i<N-1;i++){
        if (A[i] >= 0) plus.push_back(A[i]);
        else minus.push_back(A[i]);
    }
    vector<pair<int,int>> ans;
    for (int i=0;i<plus.size()-1;i++){
        int x, y;
        x = minus[minus.size()-1];
        y = plus[i];
        ans.emplace_back(x,y);
        minus[minus.size()-1] = x - y;
    }
    for (int i=0;i<minus.size();i++){
        int x, y;
        x = plus[plus.size()-1];
        y = minus[i];
        ans.emplace_back(x,y);
        plus[plus.size()-1] = x - y;
    }
    cout << plus[plus.size()-1] << "\n";
    for (int i=0;i<ans.size();i++){
        cout << ans[i].first << " " << ans[i].second << "\n";
    }
}