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;
const int N=1e5+10;
int n,m,a[N],sm,j;
int cmp(int x,int y)
{
	return x>y;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+n+1);m=a[n];j=1;
	if(a[1]<0&&a[n]>=0)
	{
		for(j=1;a[j]<0&&a[j+1]<0;j++)a[n]-=a[j];
	}
	//if(a[n]>=0)
	{
		//if(a[1]<0)
		//{int sum=0;
		//	for(int i=1;a[i]<0;i++)a[n]-=a[i],j=i+1;
		//}else j=1;
	}
	if(a[n]<0)sort(a+1,a+n+1,cmp);
	//for(int i=1;i<=n;i++)cout<<a[i]<<" ";cout<<endl;
	sm=a[j];
	for(int i=j+1;i<n;i++)
	sm-=a[i];
	printf("%d\n",abs(a[n]-sm));sm=a[j];
	for(int i=1;i<j;i++)
	{
		//if(a[n]>=0)
		printf("%d %d\n",m,a[i]);
		m-=a[i];
	}
	
	for(int i=j;i<n-1;i++)
	printf("%d %d\n",sm,a[i+1]),sm-=a[i+1];
	if(a[n]<sm)swap(a[n],sm);
	printf("%d %d",a[n],sm);
}
