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,m,ans,A,B,BA,a[120000];

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;++i){
		scanf("%d",a+i);
	}
	sort(a+1,a+n+1);
	ans=-a[1]+a[n]; for (int i=2;i<n;++i) ans+=abs(a[i]);
	printf("%d\n",ans);
	int x=a[1],y=a[n];
	for (int i=2;i<n;++i)
		if (a[i]<0){printf("%d %d\n",y,a[i]); y-=a[i];}
		else{printf("%d %d\n",x,a[i]); x-=a[i];}
	printf("%d %d\n",y,x);
	
	return 0;
}
			