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<cstring>
#include<algorithm>
#include<functional>
using namespace std;
const int MAXN=100005;

int N,cnt1,cnt2;
long long A[MAXN],B[MAXN];

int main()
{
	scanf("%d",&N);
	for(int i=1,x;i<=N;i++)
	{
		scanf("%d",&x);
		if(x>0)
			A[++cnt1]=x;
		else
			B[++cnt2]=x;
	}
	sort(A+1,A+cnt1+1,greater<long long>());
	sort(B+1,B+cnt2+1);
	if(cnt1==0)
		A[++cnt1]=B[--cnt2];
	if(cnt2==0)
		B[++cnt2]=A[cnt1--];
	long long ans=0;
	for(int i=1;i<=cnt1;i++)
		ans+=A[i];
	for(int i=1;i<=cnt2;i++)
		ans-=B[i];
	printf("%lld\n",ans);
	for(int i=1;i<N;i++)
	{
		if(cnt1>cnt2)
		{
			printf("%lld %lld\n",B[cnt2],A[cnt1]);
			B[cnt2]=B[cnt2]-A[cnt1--];
		}
		else
		{
			printf("%lld %lld\n",A[cnt1],B[cnt2]);
			A[cnt1]=A[cnt1]-B[cnt2--];
		}
	}
	return 0;
}
