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>
#pragma O3
#define int long long
using namespace std;
vector<int> r;
signed main(){
//	freopen(".inp","r",stdin);
//	freopen(".out","w",stdout);
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int n,u;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>u;
		r.push_back(u);
	}
	sort(r.begin(),r.end());
	int en=r[n-1],st=r[0];
	for(int i=1;i<n-1;i++){
		if(r[i]<0) en-=r[i];
		else st-=r[i]; 
	}
	cout<<en-st<<endl;
	for(int i=1;i<n-1;i++){
		if(r[i]<0){
			cout<<r[n-1]<<" "<<r[i]<<endl;
			r[n-1]-=r[i];
		}
		else{
			cout<<r[0]<<" "<<r[i]<<endl;
			r[0]-=r[i];
		}
	}
	cout<<r[n-1]<<" "<<r[0];
}