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: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,a,b) for(int i=(a);i<(b);i++)
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second

using vint=vector<int>;
using pint=pair<int,int>;
using vpint=vector<pint>;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<class A,class B>
ostream& operator<<(ostream& ost,const pair<A,B>&p){
	ost<<"{"<<p.first<<","<<p.second<<"}";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost,const vector<T>&v){
	ost<<"{";
	for(int i=0;i<v.size();i++){
		if(i)ost<<",";
		ost<<v[i];
	}
	ost<<"}";
	return ost;
}

signed main(){
	int N;
	cin>>N;
	vint A(N);rep(i,N)cin>>A[i];

	int m=min_element(all(A))-A.begin();
	int M=max_element(all(A))-A.begin();

	vpint ans;
	if(A[M]<=0){
		rep(i,N){
			if(i==M)continue;
			ans.eb(A[M],A[i]);
			A[M]-=A[i];
		}
	}
	else if(A[m]>=0){
		int last=N-1;
		if(last==m)last--;
		rep(i,N){
			if(i==last){
				ans.eb(A[last],A[m]);
				A[last]-=A[m];
				continue;
			}
			if(i==m)continue;
			ans.eb(A[m],A[i]);
			A[m]-=A[i];
		}
	}
	else{
		rep(i,N){
			if(i==M||i==m)continue;
			if(A[i]>=0){
				ans.eb(A[m],A[i]);
				A[m]-=A[i];
			}
		}
		rep(i,N){
			if(i==M)continue;
			if(A[i]>=0)continue;
			ans.eb(A[M],A[i]);
			A[M]-=A[i];
		}
	}

	cout<<ans.back().fi-ans.back().se<<endl;
	rep(i,N-1)cout<<ans[i].fi<<" "<<ans[i].se<<endl;
	return 0;
}