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>
using namespace std;

int main() {
	int N;
	cin >> N;
	vector<int> A(N);
	for(auto &x: A) cin >> x;

	sort(A.begin(), A.end());

	long long M = 0;
	for(int i = 0; i < N ; i++) {
		if(i == 0) M -= A[i];
		else if(i == N - 1) M += A[i];
		else if(A[i] < 0) M -= A[i];
		else M += A[i];
	}
	cout << M << endl;
	int P = 1, Q = 1;
	for(int i = 1; i < N - 1; i++) {
		if(A[i] < 0) P++;
		else Q++;
	}

	long long a = A[0], b = A[N - 1];

	while(Q > 1) {
		cout << a << " " << A[N - Q] << endl;
		a -= A[N - Q];
		Q--;
	}

	while(P > 1) {
		cout << b << " " << A[P - 1] << endl;
		b -= A[P - 1];
		P--;
	}

	cout << b << " " << a << endl;
}
