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 <algorithm>
#include <vector>
#include <cmath>
#include <map>
#include <utility>
using namespace std;

pair<int, int> make_pair(int x, int y){
	pair<int, int> tmp;
	tmp.first = x;
	tmp.second = y;
	return tmp;
}

int main(){
	int N;
	cin >> N;
	vector<long> A(N);
	for(int i=0;i<N;i++){
		cin >> A[i];
	}
	sort(A.begin(), A.end());
	int left = A[N-1], right = A[0];
	vector<pair<int, int>> ans(N-1);
	for(int i=1;i<N-1;i++){
		if(A[i] < 0){
			ans[i-1] = make_pair(left, A[i]);
			left -= A[i];
		}
		else{
			ans[i-1] = make_pair(right, A[i]);
			right -= A[i];
		}
	}
	ans[N-2] = make_pair(left, right);
	cout << left - right << endl;
	for(int i=0;i<N-1;i++){
		cout << ans[i].first << " " << ans[i].second << endl;
	}
}