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 N 100000
int a[N+1];
int main(){
	int n,sum=0,po=0,i;scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",a+i),sum+=abs(a[i]),po+=a[i]>0;
	sort(a+1,a+n+1);
	if(po==n){
		printf("%d\n",sum-2*a[1]);
		for(i=2;i<n;i++)printf("%d %d\n",a[1],a[i]),a[1]-=a[i];
		printf("%d %d\n",a[n],a[1]);
	}
	else if(!po){
		printf("%d\n",sum+2*a[n]);
		for(i=n-1;i;i--)printf("%d %d\n",a[n],a[i]),a[n]-=a[i];
	}
	else{
		printf("%d\n",sum);
		for(i=n-po+1;i<n;i++)printf("%d %d\n",a[1],a[i]),a[1]-=a[i];
		printf("%d %d\n",a[n],a[1]);a[n]-=a[1];
		for(i=2;i<=n-po;i++)printf("%d %d\n",a[n],a[i]),a[n]-=a[i];
	}
	return 0;
}
/*1
3
1 -1 2
*/
/*2
3
1 1 1
*/