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 main(){
    int N;
    cin >> N;
    vector<string> C(N-1);
    vector<int> A(N);
    for(int i{}; i<N; i++){
        cin >> A.at(i);
    }
    for(int i{}; i<N-2; i++){
        sort(A.begin(), A.end());
        C.at(i) = to_string(A.at(0)) + " " + to_string(A.at(N-i-1));
        int B = A.at(0) - A.at(N-i-1);
        A.pop_back();
        for(int j{}; j<N-i-2; j++){
            A.at(j) = A.at(j+1);
        }
        A.at(N-i-2) = B;
    }
    sort(A.begin(), A.end());
    C.at(N-2) = to_string(A.at(1)) + " " + to_string(A.at(0));
    int B = A.at(1) - A.at(0);
    cout << B << endl;
    for(int i{}; i<N-1; i++){
        cout << C.at(i) << endl;
    }
}