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<algorithm>
#include<vector>
using namespace std;
int N;
long A[1<<17];
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)cin>>A[i];
	sort(A,A+N);
	reverse(A,A+N);
	vector<pair<long,long> >S;
	for(int i=1;i<N-1;i++)
	{
		if(A[i]>=0)
		{
			S.push_back(make_pair(A[N-1],A[i]));
			A[N-1]-=A[i];
		}
		else
		{
			S.push_back(make_pair(A[0],A[i]));
			A[0]-=A[i];
		}
	}
	S.push_back(make_pair(A[0],A[N-1]));
	A[0]-=A[N-1];
	cout<<A[0]<<endl;
	for(pair<int,int>p:S)
	{
		cout<<p.first<<" "<<p.second<<endl;
	}
}