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;
#define LL long long
#define PB push_back
#define MP make_pair

int N;
LL A[100000];

int main(){
	cin >> N;
	for(int i=0; i<N; i++){
		cin >> A[i];
	}
	sort(A, A+N);
	LL M = A[N-1] - A[0];
	for(int i=1; i<N-1; i++) M += abs(A[i]);
	cout << M << endl;
	LL small = A[0];
	LL big = A[N-1];
	for(int i=1; i<N-1; i++){
		if(A[i] < 0){
			cout << big << " " << A[i] << endl;
			big -= A[i];
		}
		else{
			cout << small << " " << A[i] << endl;
			small -= A[i];
		}
	}
	cout << big << " " << small << endl;
}
