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 <cstdio>
#include <algorithm>
#include <vector>
#define N 100005
using namespace std;

int n, w1, w2, a[N];
int main()
{
	int i;
	cin >> n;
	for (i = 0; i < n; i++) {
		scanf ("%d", &a[i]);
	}
	sort(a, a + n);
	w1 = a[n - 1];
	w2 = a[0];
	for (i = 1; i < n - 1; i++) {
		if (a[i] > 0) w2 -= a[i];
		else w1 -= a[i];
	}
	cout << w1 - w2 << endl;
	w1 = a[n - 1];
	w2 = a[0];
	for (i = 1; i < n - 1; i++) {
		if (a[i] > 0) break;
		printf("%d %d\n", w1, a[i]);
		w1 -= a[i];
	}
	for (; i < n - 1; i++) {
		printf("%d %d\n", w2, a[i]);
		w2 -= a[i];
	}
	printf("%d %d\n", w1, w2);
    return 0;
}
