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;
typedef long long LL;
LL a[1000000];
int main(){
	LL n;
	cin>>n;
	LL S=0;
	for(LL i=1;i<=n;i++){
		cin>>a[i];
		S+=a[i];
	} 
	sort(a+1,a+n+1); 
	S=-a[1]+a[n];
	for(LL i=2;i<n;i++) S+=abs(a[i]);
	cout<<S<<endl;
	vector<LL> X,Y;
	X.push_back(a[n]);
	Y.push_back(a[1]);
	for(LL i=2;i<n;i++){
		if(a[i]>=0) X.push_back(a[i]);
		else Y.push_back(a[i]);
	} 
	LL now=X[0];
	LL now1;
	for(LL i=0;i<(LL)Y.size()-1;i++){
		 cout<<now<<' '<<Y[i]<<endl;
		 now=now-Y[i];
	}
	now1=Y.back();
	for(LL i=1;i<(LL)X.size();i++){
		cout<<now1<<' '<<X[i]<<endl;
		now1=now1-X[i];
	}
	cout<<now<<' '<<now1<<endl;
	return 0;
}
