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;

int main() {
	long long n,z;
	cin>>n;
	long long a[n];
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	sort(a,a+n);
	long long o=a[0];
	vector<pair<long long,long long>> v;
	if(a[n-1]<0){
		for(int i=1;i<n;i++){
		v.push_back(make_pair(o,a[i]));
		o-=a[i];
		}
		cout<<o<<endl;
		for(int i=0;i<v.size();i++){
			cout<<v[i].first<<" "<<v[i].second<<endl;
		}
	}
	else if(a[0]>=0){
		z=lower_bound(a,a+n,0)-a;
		o=a[z];
		for(int i=z+1;i<n-1;i++){
		v.push_back(make_pair(o,a[i]));
			o-=a[i];
		}
		v.push_back(make_pair(a[n-1],o));
			o=a[n-1]-o;
			for(int i=z-1;i>=0;i--)
			{
		v.push_back(make_pair(o,a[i]));
			o-=a[i];	
			}
			
		cout<<o<<endl;
		for(int i=0;i<v.size();i++){
			cout<<v[i].first<<" "<<v[i].second<<endl;
		}
	}
	else{
		z=lower_bound(a,a+n,0)-a;
		for(int i=z;i<n-1;i++){
		v.push_back(make_pair(a[z-1],a[i]));
			a[z-1]=a[z-1]-a[i];
		}
		
		for(int i=z-1;i>=0;i--){
		v.push_back(make_pair(a[n-1],a[i]));
		
			a[n-1]=a[n-1]-a[i];
		}
		cout<<a[n-1]<<endl;
		for(int i=0;i<v.size();i++){
			cout<<v[i].first<<" "<<v[i].second<<endl;
		}
		
	}
	
	return 0;
}