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: Accepted)

/* Blog: https://misteer.hatenablog.com/entry/diverta2019-2-C */

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

using namespace std;

int main() {
    /* ----- 入力 ----- */
    int N;
    cin >> N;
    vector<int> A(N);
    for (auto& a : A) cin >> a;
    sort(A.begin(), A.end());

    /* ----- 最大値を求める ----- */
    int sum = 0;
    for (auto& a : A) sum += abs(a);
    if (A.front() > 0) sum -= A.front() * 2;  // 全て正の場合
    if (A.back() < 0) sum += A.back() * 2;    // 全て負の場合
    cout << sum << endl;

    /* ----- 操作列の構成 ----- */
    int acc = A.front();

    // 正の要素を引く
    for (int i = 1; i < N - 1; ++i) {
        if (A[i] >= 0) {
            cout << acc << " " << A[i] << endl;
            acc -= A[i];
        }
    }

    // 正負逆転
    cout << A.back() << " " << acc << endl;
    acc = A.back() - acc;

    // 負の要素を引く
    for (int i = 1; i < N - 1; ++i) {
        if (A[i] < 0) {
            cout << acc << " " << A[i] << endl;
            acc -= A[i];
        }
    }
    return 0;
}