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<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>
using namespace std;
int n,a[101010],ans;
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	sort(a,a+n);
	ans=a[n-1]-a[0];
	for(int i=1;i<n-1;i++)
	{
		ans+=abs(a[i]);
	}
	cout<<ans<<endl;
	for(int i=1;i<n-1;i++)
	{
		if(a[i]<0)
		{
			cout<<a[n-1]<<' '<<a[i]<<endl;
			a[n-1]-=a[i];
		}
		else
		{
			cout<<a[0]<<' '<<a[i]<<endl;
			a[0]-=a[i];
		}
	}
	cout<<a[n-1]<<' '<<a[0]<<endl;
	return 0;
}