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 100005
int a[N],mx,n,mn,pm,pn;
int siz;pair<int ,int >ans[N];
int main()
{
	scanf("%d",&n);mx=-0x3f3f3f3f,mn=0x3f3f3f3f;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),(mx<a[i]?(mx=a[i],pm=i):0),(mn>a[i]?(mn=a[i],pn=i):0);
	if(pm==pn)
	{
		if(n==1)return printf("%d\n",mx),0;
		pn++;
	}
	for(int i=1;i<=n;i++)
		if(i!=pm&&i!=pn)
			if(a[i]<0)ans[++siz]=make_pair(mx,a[i]),mx-=a[i];
			else ans[++siz]=make_pair(mn,a[i]),mn-=a[i];
	ans[++siz]=make_pair(mx,mn);
	printf("%d\n",mx-mn);
	for(int i=1;i<=siz;i++)printf("%d %d\n",ans[i].first,ans[i].second);
}