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 rep(i,x,y) for(int i=x;i<y;i++)
#define range(a) a.begin(),a.end()

int main(){
	int N;cin>>N;
	vector<int> A(N);
	int count=0,sum=0;
	rep(i,0,N){
		cin>>A[i];
		if(A[i]<=0){
			count++;
			sum-=A[i];
		}else{
			sum+=A[i];
		}
	}sort(range(A));
	if(count==0){
		count=1;
		sum-=2*A[0];
	}else if(count==N){
		count=N-1;
		sum+=2*A[N-1];
	}reverse(A.begin()+1,A.end());
	cout<<sum<<endl;
	rep(i,1,N){
		if(i<=N-count-1){
		    cout<<A[0]<<" "<<A[i]<<endl;
		    A[0]-=A[i];
		}else if(i==N-count){
			cout<<A[i]<<" "<<A[0]<<endl;
			A[0]=A[i]-A[0];
		}else{
			cout<<A[0]<<" "<<A[i]<<endl;
			A[0]-=A[i];
		}
	}
}
