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 n,ans;
int a[100005];
vector<pair<int,int> > res;
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	sort(a,a+n);
	for(int i=n-2;i>=1;i--){
		if(a[i]<0) break;
		res.push_back(make_pair(a[0],a[i]));
		a[0]-=a[i];
	}
	for(int i=1;i<=n-2;i++){
		if(a[i]>=0) break;
		res.push_back(make_pair(a[n-1],a[i]));
		a[n-1]-=a[i];
	}
	ans=a[n-1]-a[0];
	res.push_back(make_pair(a[n-1],a[0]));
	printf("%d\n",ans);
	for(int i=0;i<res.size();i++){
		printf("%d %d\n",res[i].first,res[i].second);
	}
}