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;
#define nn 200008
int n;int val[nn];
vector<int> ex,ey;

void sub(int x,int y){
	ex.push_back(val[x]),ey.push_back(val[y]);
	val[x]-=val[y];
}
int main()
{
	cin>>n;for(int i=1;i<=n;i++) scanf("%d",&val[i]);
	sort(val+1,val+n+1);
	if(val[1]>0){
		for(int i=2;i<n;i++) sub(1,i);
		sub(n,1);
	}
	else if(val[n]<0){
		for(int i=1;i<n;i++) sub(n,i);
		
	}
	else{
		for(int i=2;i<n;i++){
			if(val[i]>0){
				sub(1,i);
			}
			else{
				sub(n,i);
			}
		}
		sub(n,1);
	}
	printf("%d\n",val[n]);
	
	for(int i=0;i<ex.size();i++) printf("%d %d\n",ex[i],ey[i]);
	return 0;
}