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;
int main(){
	int N; cin>>N;
	vector<int> A(N);
	for (int i=0; i<N; i++)
		cin>>A[i];
	sort(A.begin(),A.end());
	int M=A.back()-A.front();
	for (int i=1; i<N-1; i++)
		M+=abs(A[i]);
	cout<<M<<endl;
	int x=A.front(), y=A.back();
	for (int i=1; i<N-1; i++) {
		if (A[i]<0) cout<<y<<" "<<A[i], y-=A[i];
		else cout<<x<<" "<<A[i], x-=A[i];
		cout<<"\n";
	}
	cout<<y<<" "<<x<<"\n";
	return 0;
}