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+5;
int n,a[N];
int main(){
//	freopen("01.txt","r",stdin);
	cin>>n;
	for(int i=1;i<=n;++i)
		cin>>a[i];
	sort(a+1,a+n+1);
	int ans=0,tmp=0;
	a[0]=-1e9;
	for(int i=1;i<=n;++i)
		if(a[i]>=0 && a[i-1]<=0){
			tmp=i;break;
		}
	if(tmp==0){
		int ans=a[n];
		for(int i=1;i<n;++i)
			ans-=a[i];
		cout<<ans<<endl;
		int res=a[n];
		for(int i=1;i<n;++i)
			cout<<res<<" "<<a[i]<<endl,res-=a[i];
		return 0;
	}
	tmp=max(tmp,2);
	for(int i=1;i<tmp;++i)
		ans-=a[i];
	for(int i=tmp;i<=n;++i)
		ans+=a[i];
	cout<<ans<<endl;
	int res=a[1];
	for(int i=tmp+1;i<=n;++i)
		cout<<res<<" "<<a[i]<<endl,
		res-=a[i];
	cout<<a[tmp]<<" "<<res<<endl;
	res=a[tmp]-res;
	for(int i=2;i<tmp;++i)
		cout<<res<<" "<<a[i]<<endl,
		res-=a[i];
	return 0;
}