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>
#define ll long long
using namespace std;
bool c1(const int &a,const int &b){return a<b;}
int n,a[100010],su;
int main(){
	cin>>n;
	for (int i=1;i<=n;i++){cin>>a[i];su+=abs(a[i]);}
	sort(a+1,a+n+1,c1);
	if (a[1]>0){
		cout<<su-a[1]-a[1]<<endl;
		for (int i=2;i<n;i++){cout<<a[1]<<' '<<a[i]<<endl;a[1]-=a[i];}
		cout<<a[n]<<' '<<a[1]<<endl;
	}else{
		if (a[n]<0){
			cout<<su+a[n]+a[n]<<endl;
			for (int i=1;i<n;i++){cout<<a[n]<<' '<<a[i]<<endl;a[n]-=a[i];}
		}else{
			cout<<su<<endl;
			int l=1,r=n;
			while (l+1<r){
				if (a[l+1]<0){cout<<a[r]<<' '<<a[l]<<endl;a[r]-=a[l];l++;}else{cout<<a[l]<<' '<<a[r]<<endl;a[l]-=a[r];r--;}
			}
			cout<<a[l+1]<<' '<<a[l]<<endl;
		}
	}
	return 0;
}