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;



int main(){
	int N;
	cin>>N;

	deque<long long> A(N);
	for(int i=0;i<N;i++)cin>>A[i];
	
	sort(A.begin(),A.end());
	long long ans = 0;
	vector<pair<long long,long long>> B(0);
	for(int i=0;i<N-2;i++){
		long long a = A.front();
		long long b = A.back();
		A.pop_back();
		A.pop_front();
		if(abs(a-b-A.back())>=abs(b-a-A.front())){
			A.push_front(a-b);
			B.push_back(make_pair(a,b));
		}
		else{
			A.push_back(b-a);
			B.push_back(make_pair(b,a));
		}
	}
	B.push_back(make_pair(A[1],A[0]));
	cout<<A[1]-A[0]<<endl;
	
	for(int i=0;i<N-1;i++){
		cout<<B[i].first<<' '<<B[i].second<<endl;
	}
	
		
	
	
	
	return 0;
}