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)

use std::fs;
use std::io::{stdin, Read};

macro_rules! parse {
    ($it: ident) => {
        $it.next().unwrap().parse().unwrap()
    };
    ($it: ident,2) => {
        (parse!($it), parse!($it))
    };
    ($it: ident,3) => {
        (parse!($it), parse!($it), parse!($it))
    };
    ($it: ident,4) => {
        (parse!($it), parse!($it), parse!($it), parse!($it))
    };
    ($it: ident, $n: expr) => {
        (&mut $it).take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    let n: usize = parse!(it);
    let mut a: Vec<isize> = parse!(it, n);
    a.sort();
    let np = a.iter().filter(|&&x| x >= 0).count();
    let nm = n - np;
    let mut stack = vec![];
    if nm == 0 {
        let mut lhs = a[0];
        for i in 2..n {
            stack.push((lhs, a[i]));
            lhs -= a[i];
        }
        stack.push((a[1], lhs));
        let ret = a[1] - lhs;
        printans(stack);
    } else if np == 0 {
        let mut lhs = a[n - 1];
        for i in 0..(n - 1) {
            stack.push((lhs, a[i]));
            lhs -= a[i];
        }
        printans(stack);
    } else {
        let mut lhs = a[0];
        for i in (nm + 1)..n {
            stack.push((lhs, a[i]));
            lhs -= a[i];
        }
        stack.push((a[nm], lhs));
        lhs = a[nm] - lhs;
        for i in 1..nm {
            stack.push((lhs, a[i]));
            lhs -= a[i];
        }
        printans(stack);
    }
}
fn printans(stack: Vec<(isize, isize)>) {
    let &(a, b) = stack.last().unwrap();
    println!("{}", a - b);
    for (i, j) in stack {
        println!("{} {}", i, j);
    }
}

fn test() {
    let mut f = fs::File::open("foo.txt").unwrap();
    let mut s = String::new();
    f.read_to_string(&mut s).unwrap();
    solve(&s);
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_main() {
        test();
    }
    #[test]
    fn test_input() {
        let s = "
        3
        1 -1 2
        ";
        solve(s);
    }
}