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;

void solve(void)
{
	int n;
	cin >> n;
	vector<int> xs(n), as, bs;
	int pc = 0, nc = 0;
	for (int& x : xs)
	{
		cin >> x;
		if (x>=0) pc++;
		else nc++;
	}
	sort(xs.begin(), xs.end());
	if (pc==0)
	{
		nc--;
		pc++;
	}
	else if(nc==0)
	{
		pc--;
		nc++;
	}
	int res = xs[0], res2 = xs[n-1];
	for (int i = nc; i < n-1; i++)
	{
		as.push_back(res);
		bs.push_back(xs[i]);
		res -= xs[i];
	}
	for (int i = 1; i < nc; i++)
	{
		as.push_back(res2);
		bs.push_back(xs[i]);
		res2 -= xs[i];
	}
	as.push_back(res2);
	bs.push_back(res);
	res2 -= res;
	cout << res2 << '\n';
	for (int i = 0; i < n-1; i++)
	{
		cout << as[i] << ' ' << bs[i] << '\n';
	}
}

int main()
{
	solve();
	//cout << "yui(*-v・)yui" << endl;
	return 0;
}
