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: Wrong Answer)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let mut a: Vec<i64> = (0..n).map(|_| input()).collect();
    a.sort();

    if a[0] > 0 {
        let m = a.iter().sum::<i64>() - a[0] * 2;
        let mut t = a[0];
        println!("{}", m);
        for i in 1..n - 1 {
            println!("{} {}", t, a[i]);
            t -= a[i];
        }
        println!("{} {}", a[n - 1], t);
    } else if a[n - 1] < 0 {
        let m = -a.iter().sum::<i64>() + a[n - 1] * 2;
        let mut t = a[n - 1];
        println!("{}", m);
        for i in (1..n - 1).rev() {
            println!("{} {}", t, a[i]);
            t -= a[i];
        }
        println!("{} {}", t, a[0]);
    } else {
        println!("{}", a.iter().cloned().map(i64::abs).sum::<i64>());
        let mut k = 0;
        while k < n && a[k] < 0 {
            k += 1;
        }
        let mut t = a[0];
        for i in k..n - 1 {
            println!("{} {}", t, a[i]);
            t -= a[i];
        }
        println!("{} {}", a[n - 1], t);
        let mut t = a[n - 1] - t;
        for i in 1..k {
            println!("{} {}", t, a[i]);
            t -= a[i];
        }
    }
}