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 a[100000];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	int ans=0;
	ans+=(a[n-1]-a[0]);
	for(int i=1;i<=n-2;i++)
		ans+=abs(a[i]);
	cout<<ans<<'\n';
	int tmp1=a[n-1],tmp2=a[0];
	for(int i=1;i<n-1;i++)
		if(a[i]<=0){
			cout<<tmp1<<' '<<a[i]<<'\n';
			tmp1-=a[i];
		}
	for(int i=1;i<n-1;i++)
		if(a[i]>0){
			cout<<tmp2<<' '<<a[i]<<'\n';
			tmp2-=a[i];
		}
	cout<<tmp1<<' '<<tmp2<<'\n';
}
