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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <utility>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
const ll MOD = 1e9+7;

ll a[100000];
vector<P> te;
int main()
{
	int n;
	ll ans=0;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
	ll m=a[0]; //最小
	ll s=a[n-1]; //最大
	//最大を増やす
	for(int i=1;i<n-1;i++)
	{
		if(a[i]>=0)
			break;
		te.push_back(P(s,a[i]));
		s-=a[i];
	}
	//最小を減らす
	for(int i=n-2;i>0;i--)
	{
		if(a[i]<0)
			break;
		te.push_back(P(m,a[i]));
		m-=a[i];
	}
	te.push_back(P(s,m));
	ans=s-m;
	cout<<ans<<endl;
	for(int i=0;i<te.size();i++)
		cout<<te[i].first<<" "<<te[i].second<<endl;
	return 0;
}