Problem:
Score : 500 points

Problem Statement

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}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/* ****TCIITB**** */

#include<bits/stdc++.h>
#define io          ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0); 
#define ll          long long
#define pb          push_back
#define	endl		'\n'
#define pii         pair<ll int,ll int>
#define vi          vector<ll int>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll int)x.size()
#define hell        1000000007
#define rep(i,a,b)	for(ll int i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp          make_pair
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
using namespace std;

#define N  100005

int main()
{
	
	io
	ll n;
	cin>>n;
	std::vector<ll> a(n);
	ll cp=0,cn=0;
	rep(i,0,n)
	{
		cin>>a[i];
		if(a[i]>0)
			cp++;
		else
			cn++;
	}
	std::vector<pii> res;
	if(cp==n)
	{
		sort(all(a));
		ll curr=a[0];
		rep(i,1,n)
		{
			if(i!=(n-1))
			{
				res.pb({curr,a[i]});
				curr=(curr-a[i]);
			}
			else
			{
				res.pb({a[i],curr});
				curr=(a[i]-curr);
			}
		}
		cout<<curr<<endl;
		for(auto i:res)
		{
			cout<<i.F<<" "<<i.S<<endl;
		}	
		exit(0);
	}
	if(cn==n)
	{
		sort(all(a));
		reverse(all(a));
		ll curr=a[0];
		rep(i,1,n)
		{
			
				res.pb({curr,a[i]});
				curr=(curr-a[i]);
			
		}
		cout<<curr<<endl;
		for(auto i:res)
		{
			cout<<i.F<<" "<<i.S<<endl;
		}	
		exit(0);
	}
		sort(all(a));
		ll stp=0;
		rep(i,0,n)
		{
			if(a[i]>=0)
			{
				stp=i;
				break;
			}
		}		
		ll curr=a[0];
		rep(i,stp,n)
		{
			if(i!=(n-1))
				{
					res.pb({curr,a[i]});
					curr=(curr-a[i]);
				}
				else
				{
					res.pb({a[i],curr});
					curr=(-curr+a[i]);	
				}
			
		}
		rep(i,1,stp)
		{
			res.pb({curr,a[i]});
			curr=(curr-a[i]);
		}
		cout<<curr<<endl;
		for(auto i:res)
		{
			cout<<i.F<<" "<<i.S<<endl;
		}	
		exit(0);

	
	
}