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>
using namespace std;
#define mp(a,b) make_pair(a,b)
#define ff first
#define setp setprecision(12)<<fixed
#define ss second
#define fori(v) for(ll i=0; i<v; i++)
#define forj(v) for(ll j=0; j<v; j++)
#define fork(v) for(ll k=0; k<v; k++)
#define forl(v) for(ll l=0; l<v; l++)
#define fort(v) for(ll t=0; t<v; t++)
#define forz(v) for(ll z=0; z<v; z++)
#define forx(v) for(ll x=0; x<v; x++)
#define ll long long
#define double long double
#define MAX (int)(2*pow(10,5) + 10)
#define N 100000
#define pb(a) push_back(a)
// #define cout out
// #define cin in
ll inf = pow(10,9);
ll modulo = inf;
double eps = 1e-10;
ifstream in;
ofstream out;

void deal(){
	ll n ;
	cin>>n;
	vector<ll> arr(n+1);
	vector<vector<ll> > dp(n+1, vector<ll> (2,0));
	fori(n)
		cin>>arr[i+1];
	dp[1][0] = arr[1];
	dp[1][1] = arr[1];
	for(ll i = 2; i<=n; i++){
		dp[i][0] = max(arr[i] - dp[i-1][1], dp[i-1][0] - arr[i]);
		dp[i][1] = min(arr[i] - dp[i-1][0], dp[i-1][1] - arr[i]);
	}
	cout<<dp[n][0]<<endl;
	vector<pair<ll,ll> > seq;
	ll ind = 0;
	for(ll i = n; i>1; i--){
		forj(2){
			if(dp[i][ind] == arr[i] - dp[i-1][j]){
				seq.pb(mp(arr[i], dp[i-1][j]));
				break;
			}
			if(dp[i][ind] == dp[i-1][j] - arr[i]){
				seq.pb(mp(dp[i-1][j], arr[i]));
				break;
			}
		}
	}
	reverse(seq.begin(),seq.end());
	for(auto& el : seq)
		cout<<el.ff<<" "<<el.ss<<'\n';
}


int main(){
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	deal();
}