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<stdio.h>
#include<algorithm>
#include<functional>
#include<cmath>
using namespace std;
int main()
{
	int N, A[100000], wa1 = 0, wa2 = 0;
	scanf("%d", &N);
	for (int i = 0; i < N; i++) {
		scanf("%d", &A[i]);
	}
	sort(A, A + N);
	wa1 = A[0]; wa2 = A[N - 1];
	for (int i = 1; i < N - 1; i++) {
		if (A[i] <= 0) {
			wa2 -= A[i];
		}
		else {
			wa1 -= A[i];
		}
	}
	printf("%d\n", wa2 - wa1);
	for (int i = 1; i < N - 1; i++) {
		if (A[i] <= 0) {
			printf("%d %d\n", A[N-1], A[i]);
			A[N - 1] -= A[i];
		}
		else {
			printf("%d %d\n", A[0], A[i]);
			A[0] -= A[i];
		}
	}
	printf("%d %d\n", A[N - 1], A[0]);
	return 0;
}