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 <utility>
#include <algorithm>
#include <vector>
using namespace std;

int n, a[100010], p, q;
vector<pair<int,int>> op;

int main(){
	cin >> n;
	p = 0;
	for (int i = 0; i < n; ++i){
		cin >> a[i];
		if (a[i] >= 0) p++;
	}
	q = n - p;
	if (p == 0){p++; q--;}
	if (q == 0){p--; q++;}
	sort(a, a+n);
	for (int i = q; i < n-1; ++i){
		op.emplace_back(a[0], a[i]);
		a[0] -= a[i];
	}
	for (int i = 0; i < q; ++i){
		op.emplace_back(a[n-1], a[i]);
		a[n-1] -= a[i];
	}
	printf("%d\n", a[n-1]);
	for (auto &o: op) printf("%d %d\n", o.first, o.second);
}