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() {
	vector<long> m;
	int n; cin >> n;
	vector<long> arr;
    for (int i = 0; i < n; i++) {
		long x; cin >> x;
		arr.push_back(x);
	}
	sort(arr.begin(), arr.end());
	long num = arr[0];
	for (int i = 1; i < n; i++) {
		if (1<=i&&i < n - 1) {
			if (num <= 0) {
				m.push_back(num); m.push_back(arr[i]);
				num = num - arr[i];
			}
			else{
				m.push_back(arr[i]); m.push_back(num);
				num = arr[i]-num;
			}
			
		}
		else if (i == n - 1) {
			if (arr[i] < num) {
				cout << num-arr[i]  << endl;

				m.push_back(num); m.push_back(arr[i]);
			}
			else {
				cout << arr[i] - num << endl;

				m.push_back(arr[i]); m.push_back(num);
			}
			
		}
	}
	for (int i = 0; i < m.size(); i++) {
		cout << m[i] << " "; if (i % 2 == 1) {
			cout << endl;
		}
	}

	return 0;
}