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 LL = long long;

int main() {
	int n;
	scanf("%d", &n);

	deque< LL> in;

	int pos_cnt = 0;

	for (int i = 0; i < n; ++i) {
		LL x;
		scanf("%lld", &x);
		in.push_back(x);
	}
	sort(in.begin(), in.end());

	vector< pair< LL, LL>> vec;
	for (int i = 0; i < n - 1; i ++) {
		LL a = in.front();
		LL b = in.back();
		in.pop_front();
		in.pop_back();

		if (i == n - 2) {
			in.push_front(b - a);
			vec.emplace_back(b, a);
		}
		else {
			if (in.back() <= 0) {
				in.push_back(b - a);
				vec.emplace_back(b, a);
			} else {
				in.push_front(a - b);
				vec.emplace_back(a, b);
			}
		}
	}

	cout << in[0] << endl;
	for (auto& x : vec) {
		printf("%lld %lld\n", x.first, x.second);
	}
}