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 <queue>
#include <cmath>
#include <algorithm>

using namespace std;
int main()
{
    int n ;
    cin >> n ;
    priority_queue<int,vector<int>,greater<int> >q;
    int sum = 0 ;
    int min = 1e5+5 ;
    for (int i = 0 ; i < n ; i++) {
        int temp ;
        cin >> temp ;
        q.push(temp) ;
        sum += temp ;
        if (min>temp) min = temp ;
    }
    cout << sum-2*min << endl ;
    for (int i = 0 ; i < n-2 ; i++) {
        int x = q.top() ;
        q.pop();
        int y = q.top() ;
        q.pop() ;
        cout << x << " " << y << endl ;
        int temp = x - y ;
        q.push(temp) ;
    }
    int y = q.top() ;
    q.pop();
    int x = q.top() ;
    q.pop() ;
    cout << x << " " << y << endl ;
    return 0;
}