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 Rust (1.15.1) (judge verdict: Accepted)

pub fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

pub fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

pub fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

fn main() {
    let n: usize = read();
    let mut a: Vec<i32> = read_vec();
    a.sort();

    if a[0] >= 0 {
        let m = a.iter().fold(0, |sum, x| sum + x) - 2 * a[0];
        println!("{}", m);
        let mut tmp = a[0];
        for i in 1..n - 1 {
            println!("{} {}", tmp, a[i]);
            tmp -= a[i];
        }
        println!("{} {}", a[n - 1], tmp);
    } else if a[n - 1] <= 0 {
        let m = 2 * a[n - 1] - a.iter().fold(0, |sum, x| sum + x);
        println!("{}", m);
        let mut tmp = a[n - 1];
        for i in (0..n - 1).rev() {
            println!("{} {}", tmp, a[i]);
            tmp -= a[i];
        }
    } else {
        let m = a.iter().fold(0, |sum, x| sum + x.abs());
        println!("{}", m);
        let mut k = 0;
        for i in 1..n {
            if a[i] > 0 {
                k = i - 1;
                break;
            }
        }
        let mut tmp1 = a[0];
        for i in k + 1..n - 1 {
            println!("{} {}", tmp1, a[i]);
            tmp1 -= a[i];
        }
        let mut tmp2 = a[n - 1];
        for i in 1..k + 1 {
            println!("{} {}", tmp2, a[i]);
            tmp2 -= a[i];
        }
        println!("{} {}", tmp2, tmp1);
    }
}