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 <bits/stdc++.h>
using namespace std;

int main(int argc, char const *argv[])
{
	int n;
	cin>>n;
	int a[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>a[i];
	}
	sort(a,a+n);
	reverse(a,a+n);
	int ans=0;
	for (int i = 0; i < n; ++i)
	{
		ans+=abs(a[i]);
	}
	if (a[0]<0)
	{
		ans-=2*abs(a[0]);
	}
	else if (a[n-1]>0)
	{
		ans-=2*abs(a[n-1]);
	}
	cout<<ans<<endl;
	int now=1;
	while (a[now]>0)
	{
		cout<<a[n-1]<<" "<<a[now]<<endl;
		a[n-1]-=a[now];
		++now;
	}
	while (now<n)
	{
		cout<<a[0]<<" "<<a[now]<<endl;;
		a[0]-=a[now];
		++now;
	}
	return 0;
}
