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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
import java.io.*;
import java.util.*;

class Main {
    static Scanner scanner = new Scanner();
    static long mod = 1000000007;

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int[] a = new int[n];
        Integer[] b = new Integer[n];
        boolean posi = false, nega = false, zero = false;
        long sum = 0;
        for (int i = 0; i < n; i++) {
            sum += a[i] = Math.abs(b[i] = scanner.nextInt());
            posi |= b[i] > 0;
            nega |= b[i] < 0;
            zero |= b[i] == 0;
        }
        Arrays.sort(a);
        if (posi && nega) {
            System.out.println(sum);
            Deque<Integer> posiDeque = new ArrayDeque<>();
            Deque<Integer> negaDeque = new ArrayDeque<>();
            Arrays.sort(b, Comparator.comparingInt(Math::abs));
            for (int i = 0; i < n; i++) {
                if (b[i] > 0) {
                    posiDeque.add(b[i]);
                } else {
                    negaDeque.add(b[i]);
                }
            }
            while (posiDeque.size() >= 2) {
                int i = posiDeque.poll();
                int j = negaDeque.poll();
                System.out.println(j + " " + i);
                negaDeque.add(j - i);
            }
            while (!negaDeque.isEmpty()) {
                int i = posiDeque.poll();
                int j = negaDeque.poll();
                System.out.println(i + " " + j);
                posiDeque.add(i - j);
            }
        } else {
            Arrays.sort(b, Comparator.comparingInt(Math::abs));
            System.out.println(zero ? sum : sum - a[0] * 2);
            for (int i = 1; i < n - 1; i++) {
                System.out.println(b[i - 1] + " " + b[i]);
                b[i] = b[i - 1] - b[i];
            }
            if (posi) {
                System.out.println(b[n - 1] + " " + b[n - 2]);
            } else {
                System.out.println(b[n - 2] + " " + b[n - 1]);
            }
        }
    }

    static class Scanner {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 1 << 15);
        StringTokenizer tokenizer;

        String next() {
            try {
                while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                    tokenizer = new StringTokenizer(reader.readLine());
                }
            } catch (IOException ignored) {
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}