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 <string>
#include <vector>
#include <algorithm>

using namespace std;

int main()

{
	
	int n, i, j, k;
	cin >> n;
	
	int arr[n];
	
	
	int ans[2*n];
	
	for (i = 0; i < n; ++i)
	{
		cin >> arr[i];
	}
	
	sort(arr, arr+n);
	int ii = 0;
	i = n-1;
	j = 0;
	while ((i > 0) && (arr[i-1] > 0))
	{
		//cout << arr[j] << " " <<  arr[i] << '\n';
		ans[ii++] = arr[j];
		ans[ii++] = arr[i];
		arr[j] = arr[j] - arr[i];
		--i;
	}
	while (j < i)
	{
		//cout << arr[i] << " " << arr[j] << '\n';
		ans[ii++] = arr[i];
		ans[ii++] = arr[j];
		arr[i] = arr[i] - arr[j];
		++j;
	}
	
	cout << arr[i] << endl;
	
	for(i = 0; i < ii; i+=2)
	{
		cout << ans[i] << " " << ans[i+1] << endl;
	}
	
	
	return 0;
}