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 <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <algorithm>
using namespace std;


int main()
{
	vector<int> x, y;
	vector<int> memo;

	int n;
	cin >> n;
	for(int i=0;i<n;i++){
		int a;
		cin >> a;
		memo.push_back(a);
	}
	sort(memo.begin(), memo.end());

	int l = memo[0];
	int r = memo[memo.size()-1];

	for(int i=1;i<n-1;i++){
		if(memo[i] >= 0){
			x.push_back(l);
			y.push_back(memo[i]);
			l = l - memo[i];
		}else{
			x.push_back(r);
			y.push_back(memo[i]);
			r = r - memo[i];
		}
	}
	x.push_back(r);
	y.push_back(l);

	cout << r - l << endl;
	for(int i=0;i<n-1;i++){
		cout << x[i] << " " << y[i] << endl;
	}
	return 0;
}