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 <string>
#include <algorithm>
#include <vector> 
#include <math.h>
#include <iostream>
using namespace std;
long long mod = 1000000000 + 7;


int main()
{
	int n;
	cin >> n;
	vector<int> num;
	for (int i = 0; i < n; i++) {
		int nin;
		cin >> nin;
		num.push_back(nin);
	}
	sort(num.begin(), num.end());
	long long ans = 0;
	
	int a = num[n - 1];
	int b = num[0];
	for (int i = 1; i < n - 1; i++) {
		if (num[i] >= 0) {
			ans = ans + num[i];
		}
		else {
			ans = ans - num[i];
		}
	}
	
	ans = ans + a - b;
	cout << ans << endl;


	for (int i = 1; i < n - 1; i++) {
		if (num[i] >= 0) {
			cout << b << " " << num[i] << endl;
			b =  b - num[i];
		}
		else {
			cout << a << " " << num[i] << endl;
			a = a - num[i];
		}
	}
	cout << a << " " << b << endl;
	
	return 0;

}
