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)

#[allow(dead_code)]
fn main() {
    let n = i::u();
    let mut a = i::iv(n);
    a.sort();
    let mut max = a[n - 1];
    let mut min = a[0];
    let mut ans = Vec::new();
    for i in 1..n - 1 {
        if a[i] < 0 {
            ans.push(format!("{} {}", max, a[i]));
            max = max - a[i];
        } else {
            ans.push(format!("{} {}", min, a[i]));
            min = min - a[i];
        }
    }
    println!("{}", max - min);
    for i in ans {
        println!("{}", i);
    }
    println!("{} {}", max, min);
}

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::io::*;
#[allow(unused_imports)]
use std::num::*;
#[allow(unused_imports)]
use std::str::*;

#[allow(dead_code)]
mod i {
    use super::*;

    pub fn read<T: FromStr>() -> T {
        stdin()
            .bytes()
            .map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>()
            .parse::<T>()
            .ok()
            .unwrap()
    }

    pub fn str() -> String {
        read()
    }

    pub fn s() -> Vec<char> {
        str().chars().collect()
    }

    pub fn i() -> i64 {
        read()
    }

    pub fn u() -> usize {
        read()
    }

    pub fn u2() -> (usize, usize) {
        (read(), read())
    }

    pub fn u3() -> (usize, usize, usize) {
        (read(), read(), read())
    }

    pub fn f() -> f64 {
        read()
    }

    pub fn c() -> char {
        read::<String>().pop().unwrap()
    }

    pub fn iv(n: usize) -> Vec<i64> {
        (0..n).map(|_| i()).collect()
    }

    pub fn iv2(n: usize) -> Vec<(i64, i64)> {
        (0..n).map(|_| iv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv(n: usize) -> Vec<usize> {
        (0..n).map(|_| u()).collect()
    }

    pub fn uv2(n: usize) -> Vec<(usize, usize)> {
        (0..n).map(|_| uv(2)).map(|a| (a[0], a[1])).collect()
    }

    pub fn uv3(n: usize) -> Vec<(usize, usize, usize)> {
        (0..n).map(|_| uv(3)).map(|a| (a[0], a[1], a[2])).collect()
    }

    pub fn fv(n: usize) -> Vec<f64> {
        (0..n).map(|_| f()).collect()
    }

    pub fn cmap(h: usize) -> Vec<Vec<char>> {
        (0..h).map(|_| s()).collect()
    }
}