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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

void swap(int *a,int *b){

    int tmp = *a;

    if(*a > *b){
        *a = *b;
        *b = tmp;
    }
}

int main(){

    int N;
    scanf("%d", &N);

    int A[N];
    for(int i = 0; i < N; i++){
        scanf("%d", &A[i]);
    }

    for (int i = N-1; i > 0; i--){
        for(int j = 0; j < i; j++){
            swap(&A[j], &A[j + 1]);
        }
    }

    int max_sum = 0;

    for (int i = 0; i < N; i++){

        if(i < N/2){
            max_sum -= A[i];
        }else{
            max_sum += A[i];
        }
    }

    printf("%d\n", max_sum);

    if(N/2 == 0){

        int temp = A[0];

        for (int i = 0; i < N-1; i++){

            if(i/2 == 0){

                printf("%d %d\n", A[N - (i / 2) - 1], temp);

                temp = A[N - (i / 2) - 1] - temp;

            }else if(i/2 == 1){

                printf("%d %d\n", A[(i + 1) / 2], temp);

                temp = A[(i + 1) / 2] - temp;

            }
        }

    }else{

        int temp = A[N - 1];

        for (int i = 0; i < N-1; i++){

            if(i/2 == 1){

                printf("%d %d\n", A[N - (i / 2) - 1], temp);

                temp = A[N - ((i - 1))/ 2] - temp;

            }else if(i/2 == 0){

                printf("%d %d\n", A[(i + 1) / 2], temp);

                temp = A[i/2] - temp;

            }
        }
    }

    return 0;
}