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 <algorithm>
using namespace std;

const int MAXN = 100010;
int a[MAXN];
vector<int> pos, neg;
int main() {
	int n;
	cin>>n;
	for (int i=0; i<n; i++)
		cin>>a[i];
	sort(a, a+n);

	int ans = 0;
	pos.push_back(a[n-1]);
	neg.push_back(a[0]);
	ans = a[n-1] - a[0];
	for (int i=1; i<n-1; i++)
		if (a[i] < 0)
			neg.push_back(a[i]), ans -= a[i];
		else
			pos.push_back(a[i]), ans += a[i];
	cout<<ans<<endl;

	int tmp = neg.back();
	neg.pop_back();
	while(pos.size() > 1) {
		cout<<tmp<<" "<<pos.back()<<endl;
		tmp = tmp - pos.back();
		pos.pop_back();
	}
	cout<<pos[0]<<" "<<tmp<<endl;
	tmp = pos[0] - tmp;
	while(neg.size()) {
		cout<<tmp<<" "<<neg.back()<<endl;
		tmp = tmp - neg.back();
		neg.pop_back();
	}
	return 0;
}