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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main{
    
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        
        int N = sc.nextInt();
        int[] a = new int[N];
        long sum = 0;
        int plus = 0;
        for(int i = 0; i < N; i++){
            a[i] = sc.nextInt();
            if(a[i]>=0) plus++;
            sum += Math.abs(a[i]);
        }
        sc.close();

        Arrays.sort(a);
        long ans = sum;
        if(plus == N)
            ans -= 2*a[0];
        else if(plus == N)
            ans += 2*a[N-1];
        System.out.println(ans);
        if(plus == N) plus--;
        if(plus == 0) plus++;
        int xy = a[N-1];
        for(int i = 1; i < N-plus; i++){
            System.out.println(xy + " " + a[i]);
            xy = xy - a[i];
        }
        int yx = a[0];
        for(int i = N-2; i >= N-plus; i--){
            System.out.println(yx + " " + a[i]);
            yx = yx - a[i];
        }
        System.out.println(xy + " " + yx);
    }
}