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 <malloc.h>
using namespace std;

int main() {
	int N;
	int A[100000];
	int max = -10001;
	int min = 10001;
	int count = 0, count2 = 0;
	int save[100000][2];
	cin >> N;
	for (int i = 0;i < N;i++) {
		cin >> A[i];
	}
	for (int i = 0;i < N;i++) {
		if (max < A[i]) {
			max = A[i];
			count = i;
		}
		else if (min > A[i]) {
			min = A[i];
			count2 = i;
		}
	}
	cout << count << "//" << count2 << endl;
	for (int i = 0;i < N;i++) {
		if (A[i] > 0 && (i != count && i != count2)) {
			save[i][0] = min;
			save[i][1] = A[i];
			/*cout << min << " " << A[i] << endl;*/
			min -= A[i];
		}
		if (A[i] <= 0 && (i != count && i != count2)) {
			save[i][0] = max;
			save[i][1] = A[i];
			/*cout << max << " " << A[i] << endl;*/
			max -= A[i];
		}
	}
	/*cout << max << " " << min << endl;*/
	cout << max - min << endl;
	for (int i = 0;i < N;i++) {
		cout << save[i][0] << " " << save[i][1] << endl;
	}
	cout << max << " " << min << endl;
	return 0;
}