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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(void) {
	int num, i, x, y;
	cin >> num;
	vector<int> d(num), X(num - 1), Y(num - 1);
	for (i = 0; i < num; i++)
		scanf("%d", &d[i]);
	sort(d.begin(), d.end());
	x = d[num - 1];
	y = d[0];
	i = 1;
	while (i < num - 1 && d[i] < 0) {
		X[i] = x;
		Y[i] = d[i];
		x -= d[i++];
	}
	while (i < num - 1) {
		X[i] = y;
		Y[i] = d[i];
		y -= d[i++];
	}
	printf("%d\n", x - y);
	for (i = 1; i < num - 1; i++)
		printf("%d %d\n", X[i], Y[i]);
	printf("%d %d\n", x, y);
	return 0;
}