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;
vector<long long> inp;
vector<pair<long long, long long> >anss;
signed main(){
	long long n;
	cin >> n;
	for(int i=0;i<n;i++){
		long long a;
		cin >> a;
		inp.push_back(a);
	}
	sort(inp.begin(),inp.end());
	long long mdn=inp[0];
	long long mdx=inp[inp.size()-1];
	for(int i=1;i<inp.size()-1;i++){
		if(inp[i]>=0){
			anss.push_back({mdn,inp[i]});
			mdn-=inp[i];
		}
		else{
			anss.push_back({mdx,inp[i]});
			mdx-=inp[i];
		}
	}
	cout << mdx-mdn<<endl;
	for(auto x: anss){
		cout << x.first << " " << x.second<<endl;
	}
	cout << mdx << " " << mdn;
}