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;


int main() {
	int n; cin >> n;
	vector<pair<int, int>>p;
	int ans = 0;
	vector<int>a;
	for (int i = 0; i < n; i++) {
		int b; cin >> b; a.push_back(b);
	}
	sort(a.begin(), a.end());

	int g = a[n - 1], l = a[0];
	for (int i = 1; i < n-1; i++) {
		if (a[i] >= 0) break;
		p.push_back(make_pair(g, a[i]));
		g -= a[i];
	}
	for (int i = n - 2; i > 0; i--) {
		if (a[i] < 0) break;
		p.push_back(make_pair(l, a[i]));
		l -= a[i];
	}
	p.push_back(make_pair(g, l));
	ans = g - l;

	cout << ans << endl;
	for (int i = 0; i < p.size(); i++) {
		cout << p[i].first << " " << p[i].second << endl;
	}

	return 0;
}
