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 <utility>
#include <set>
#include <algorithm>
#include <vector>
#define llint long long

using namespace std;
typedef pair<llint, llint> P;

llint n;
llint a[100005];
P p[100005];
vector<P> vec;

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	sort(a+1, a+n+1);
	reverse(a+1, a+n+1);
	
	for(int i = n-1; a[i] < 0 && i >= 2; i--){
		vec.push_back(make_pair(a[1], a[i]));
		a[1] -= a[i];
	}
	for(int i = 2; a[i] >= 0 && i <= n-1; i++){
		vec.push_back(make_pair(a[n], a[i]));
		a[n] -= a[i];
	}
	vec.push_back(make_pair(a[1], a[n]));
	cout << a[1] - a[n] << endl;
	for(int i = 0; i < vec.size(); i++) cout << vec[i].first << " " << vec[i].second << endl;
	
	return 0;
}