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 maxn=100100;
int n;
long long a[maxn];
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++) scanf("%lld",a+i);
	sort(a+1,a+n+1,greater<long long>());
	int st=n;
	for(int i=2;i<=n;i++) if(a[i]<0) {st=i;break;}
	long long ans=0;
	for(int i=1;i<st;i++) ans+=a[i];
	for(int i=st;i<=n;i++) ans-=a[i];
	cout<<ans<<endl;
	for(int i=2;i<st;i++) cout<<a[st]<<' '<<a[i]<<endl,a[st]-=a[i];
	for(int i=st;i<=n;i++) cout<<a[1]<<' '<<a[i]<<endl,a[1]-=a[i];
	return 0;
}