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 EM 100000000
using namespace std;
using LL = long long;
using P = pair<LL, LL>;
LL LINF = 1e18;
int INF = 1e9;
using vint = vector<int>;
using vLL = vector<LL>;
using vvint = vector<vector<int>>;
using vvLL = vector<vector<LL>>;

int main(){
	int N;
	cin >> N;
	vint a(N);
	for(int i = 0;i < N;i++)	cin >> a[i];
	sort(a.begin(), a.end());
	vector<P>	ans;
	int id = N-1;
	for(int i = 0;i < N;i++){
		if(a[i] > 0){
			id = i;
			break;
		}
	}
	for(int i = id+!id;i < N-1;i++){
		ans.push_back(P(a[max(0, id-1)], a[i]));
		a[max(0, id-1)] -= a[i];
	}
	for(int i = 0;i < id+!id;i++){
		ans.push_back(P(a[N-1], a[i]));
		a[N-1] -= a[i];
	}
	cout << a[N-1] << endl;
	for(auto an : ans)	cout << an.first << " " << an.second << endl;
}