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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int cmp( const void *p, const void *q ) {
    return *(int*)p - *(int*)q;
}

int main(void)
{
    int n, i, j, m, x;
    scanf("%d", &n);
    m = ((n * n) - n) / 2;
    int a[n];

    for (i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }

    x = sizeof( a )/sizeof( int );
    qsort( a, x, sizeof(int), cmp );

    int l, max = 0, flg1, flg2, s = 0, k, ans, con;
    for (i = 0; i < n - 1; i++) {
        for (j = i + 1; j < n; j++) {
            s = a[j] - a[i];
            ans = s;
            for (k = 0; k < n; k++) {
                if (k != i && k != j) {
                    s += a[k];
                }
            }
            if (s > max) {
                max = s;
                flg1 = j;
                flg2 = i;
                con = ans;
            }
        }
    }
    l = max;
    for (i = 0; i > n; i++) {
        if (i != flg1 && i != flg2) {
            l += a[i];
        }
    }

    printf("%d\n", l);
    printf("%d %d\n", a[flg2], a[flg1]);
    j = 0;
    int b[n - 2];
    for (i = 0; i < n; i++) {
        if (i != flg1 && i != flg2) {
            b[j] = a[i];
            j++;
        }
    }

    for (i = 0; i < n - 3; i++) {
        printf("%d %d\n", -con, b[i]);
        con = con + b[i];
    }

    printf("%d %d\n", b[n - 3], -con);

    return 0;

}