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"
#define in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;

int main()
{
	LL N;
	in >> N;
	std::vector<LL>A(N);
	rep(i, N) in >> A[i];

	LL border = -1;
	std::sort(A.begin(), A.end());
	rep(i, N - 1)
	{
		if (A[i] < 0 && A[i + 1] >= 0)
		{
			border = i;
			break;
		}
	}
	if (border == -1)
	{
		if (A[0] >= 0) border = 0;
		else border = N - 2;
	}

	std::vector<std::pair<LL, LL>>xy;
	for (LL i = border + 2; i < N; ++i)
	{
		xy.push_back(std::make_pair(A[border], A[i]));
		A[border] -= A[i];
	}
	rep(i, border + 1)
	{
		xy.push_back(std::make_pair(A[border + 1], A[i]));
		A[border + 1] -= A[i];
	}

	out << A[border + 1] << std::endl;
	rep(i, N - 1) out << xy[i].first << " " << xy[i].second << std::endl;
}
