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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    int n;
    cin >> n;

    vector<int> plus, minus;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        if (a > 0)plus.push_back(a);
        else minus.push_back(a);
    }
    sort(plus.begin(), plus.end()), sort(minus.begin(), minus.end());

    int ret = 0;
    vector<vector<int>> ope;
    if (plus.size() > 0 && minus.size() > 0) {
        for (int i = 1; i < minus.size(); i++) {
            ope.push_back({plus.front(), minus[i]});
            plus.front() -= minus[i];
        }
        for (int i = 1; i < plus.size(); i++) {
            ope.push_back({minus.front(), plus[i]});
            minus.front() -= plus[i];
        }
        ope.push_back({plus.front(), minus.front()});
        ret = plus.front() - minus.front();
    } else if (plus.size() > 0) {
        for (int i = 2; i < plus.size(); i++) {
            ope.push_back({plus.front(), plus[i]});
            plus.front() -= plus[i];
        }
        ope.push_back({plus[1], plus.front()});
        ret = plus[1] - plus.front();
    } else {
        for (int i = 1; i < minus.size(); i++) {
            ope.push_back({minus.front(), minus[i]});
            minus.front() -= minus[i];
        }
        ret = minus.front();
    }

    cout << ret << endl;
    for (auto x:ope)cout << x[0] << " " << x[1] << endl;
    return 0;
}