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;
const int N=1e5+10;
int n,a[N];

int main()
{
	cin>>n;
	int Max=1,Min=1;
	for (int i=1;i<=n;i++) {
		scanf("%d",&a[i]);
		if (a[i]>a[Max]) Max=i;
		if (a[i]<a[Min]) Min=i;
	}
	if (Max==Min) Max=2;
	
	long long sum=0;
	for (int i=1;i<=n;i++)
		if (i!=Max && i!=Min) sum+=abs(a[i]);
	cout<<(long long)a[Max]-a[Min]+sum<<endl;	
	
	for (int i=1;i<=n;i++)
		if (i!=Max && i!=Min && a[i]>=0) printf("%d %d\n",a[Min],a[i]),a[Min]-=a[i];
	for (int i=1;i<=n;i++)
		if (i!=Max && i!=Min && a[i]<0) printf("%d %d\n",a[Max],a[i]),a[Max]-=a[i];		
	printf("%d %d\n",a[Max],a[Min]);	
	return 0;
} 