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<cstdio>
#include<algorithm>
using namespace std;
int a[100050];
int main(){
	int n,i,p=0,ans;
	scanf("%d",&n);
	for(i=0;i<n;++i)scanf("%d",a+i);
	sort(a,a+n);
	while(p<n&&a[p]<0)++p;
	if(!p){
		ans=-a[0];
		for(i=1;i<n;++i)ans+=a[i];
		printf("%d\n",ans);
		for(i=1;i<n-1;++i){
			printf("%d %d\n",a[0],a[i]);
			a[0]-=a[i];
		}
		printf("%d %d\n",a[n-1],a[0]);
	}else if(p<n){
		ans=0;
		for(i=0;i<n;++i)ans+=abs(a[i]);
		printf("%d\n",ans);
		for(i=p;i<n-1;++i){
			printf("%d %d\n",a[0],a[i]);
			a[0]-=a[i];
		}
		for(i=0;i<p;++i){
			printf("%d %d\n",a[n-1],a[i]);
			a[n-1]-=a[i];
		}
	}else{
		ans=a[n-1];
		for(i=0;i<n-1;++i)ans-=a[i];
		printf("%d\n",ans);
		for(i=0;i<n-1;++i){
			printf("%d %d\n",a[n-1],a[i]);
			a[n-1]-=a[i];
		}
	}
	return 0;
}