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 n,a[111111],pos,ans,res;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	sort(a+1,a+n+1);
	pos=1;
	for (int i=2;i<n;i++)
	{
		if (a[i]<0) pos=i;
		else break; 
	}
	for (int i=1;i<=n;i++)
	{
		if (i<=pos) ans-=a[i];
		else ans+=a[i];
	}
	printf("%d\n",ans);
	res=a[1];
	if (pos+1!=n)
	{
		printf("%d %d\n",a[1],a[pos+1]);
		res=a[1]-a[pos+1];
		for (int i=pos+1;i<n-1;i++) 
		{
			printf("%d %d\n",res,a[i+1]);
			res-=a[i+1];
		}
	}
	printf("%d %d\n",a[n],res);
	res=a[n]-res;
	for (int i=2;i<=pos;i++)
	{
		printf("%d %d\n",res,a[i]);
		res-=a[i];
	}
	return 0;
}