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> An(N);
	vector<int> x(N-1);
	vector<int> y(N-1);
 	int loop,count=0;
	for(loop=0;loop<N;loop++){
		cin >> An[loop];
	}
	sort(An.begin(),An.end());
	for(loop=1;An[loop]<0;loop++){
		x[count]=An[N-1];
		y[count]=An[loop];
		count++;
		An[N-1]-=An[loop];
	}
	for(;loop<N-1;loop++){
		x[count]=An[0];
		y[count]=An[loop];
		count++;
		An[0]-=An[loop];
	}
	x[count]=An[N-1];
	y[count]=An[0];
	An[N-1]-=An[0];
	cout << An[N-1] << endl;
	for(loop=0;loop<N-1;loop++){
		cout << x[loop] << " " << y[loop] << endl;
	}
	return 0;
}