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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <iomanip>

using namespace std;

int const N = 55;
int const M = 1e9;

int n,i,ne,po,res,f,s;
vector<pair<int,int>> ans;

int main() {
	
	cin>>n;
	int a[n];
	for(i=0;i<n;i++) {
		cin>>a[i];
		if(a[i]<0) ne++;
		else po++;
	}
	
	sort(a,a+n);
	
	if(ne==0){
		
		f = a[0];
		for(i=1;i<n-1;i++){
			ans.push_back({f,a[i]});
			f = f-a[i];
		}
		ans.push_back({a[n-1],f});
		cout<<a[n-1]-f<<endl;
	}
	else {
		f = a[n-1];
		for(i=0;i<ne-1;i++) { ans.push_back({f,a[i]});; f = f-a[i]; }
		s = a[ne-1];
		for(i=ne;i<n-1;i++) { ans.push_back({s,a[i]});; s = s-a[i]; }
		ans.push_back({f,s});
		cout<<f-s<<endl;
	}
	
	for(i=0;i<ans.size();i++) cout<<ans[i].first<<" "<<ans[i].second<<endl;
	return 0;
}