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)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define all(a) a.begin(),a.end()
#define F first
#define S second
#define pb push_back
#define ll long long
#define vi vector<int>
#define pi pair<int,int>
#define mp make_pair

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) "FUCK YOU"
#endif

const int mod=1e9+7;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename A,typename B> istream& operator>>(istream&,pair<A,B>&);
template<typename A> istream& operator>>(istream&,vector<A>&);
void add(int&,int);
int mul(int,int);
int powz(int,int);
template<typename A> ostream& operator<<(ostream&,vector<A>&);

const int N=500006;

void solve(){
	int n;
	cin>>n;
	vector<pi> a;
	for(int i=0;i<n;i++){
		int x;
		cin>>x;
		a.pb({x,i});
	}
	int ind=0;
	ll val=0;
	vector<pi>ans;
	sort(all(a));
	debug(a);
	while(ind<n&&a[ind].F<0){
		ind++;
	}
	if(ind==0&&a[ind].F>=0){
		for(int i=0;i<n-1;i++){
			if(i==0)
			val=a[i].F;
			ans.pb({a[i+1].F,val});
			val=a[i+1].F-val;
		}
	}
	else if(ind==n){
		for(int i=0;i<n-1;i++){
			if(i==0)
			val=a[i].F;
			ans.pb({a[i+1].F,val});
			val=a[i+1].F-val;
		}
		
	}
	else{
		//~ cout<<"CASE";
		int val1=a[0].F;
		for(int i=ind;i<n-1;i++){
			debug(i);
			ans.pb({val1,a[i].F});
			val1-=a[i].F;
		}
		int val2=a[n-1].F;
		for(int i=1;i<ind;i++){
			ans.pb({val2,a[i].F});
			val2-=a[i].F;
		}
		ans.pb({val2,val1});
			val=val2-val1;
		
		
	}
	cout<<val<<"\n";
	for(auto i:ans)cout<<i.F<<' '<<i.S<<"\n";
			
}
	
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int tc=1;
	//~ cin>>tc;
	for(int _=0;_<tc;_++){
		//~ cout<<"Case #"<<_+1<<": ";
		solve();
		if(_!=tc-1)
		cout<<"\n";
	}


}

int mul(int a,int b)
{
	return ((a)*1ll*(b))%mod;
}

void add(int &a,int b)
{
	a+=b;
	if(a>=mod)a-=mod;
	if(a<0)a+=mod;
}

int powz(int a,int b)
{
	int res=1;
	while(b)
	{
		if(b&1)res=mul(res,a);
		b/=2;
		a=mul(a,a);
	}
	return res;
}

template <typename A, typename B>
istream& operator>>(istream& input,pair<A,B>& x) {
	input>>x.F>>x.S;
	return input;
}

template <typename A>
istream& operator>>(istream& input,vector<A>& x) {
	for(auto& i:x)
		input>>i;
	return input;
}

template<typename A>
ostream& operator<<(ostream& output,vector<A>& x) {
	for(auto& i:x)
	    output<<i<<' ';
	return output;	
}