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;
using lint = long long;

signed main(){
	lint N, sum = 0;
	cin >> N;
	vector<lint> a(N);
	for(lint i = 0; i < N; i++) cin >> a[i], sum += abs(a[i]);
	sort(a.begin(), a.end());
	if(a[0] > 0) sum -= a[0] * 2;
	if(a.back() < 0) sum += a.back() * 2;
	cout << sum << endl;
	lint x = a[0];
	for(lint i = 1; i < N - 1; i++){
		if(a[i] >= 0){
	 	  cout << x << " " << a[i] << endl;
		  x -= a[i];
		}
	}
	cout << a.back() << " " << x << endl;
	x = a.back() - x;
	for(lint i = 1; i < N - 1; i++){
		if(a[i] < 0){
			cout << x << " " << a[i] << endl;
			x -= a[i];
		}
	}
}