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 (1.42.0) (judge verdict: Wrong Answer)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use ascii::AsciiChar;
use ascii::AsciiStr;
use itertools::Itertools as _;
use permutohedron::LexicalPermutation as _;
use petgraph::algo::dijkstra;
use petgraph::algo::min_spanning_tree;
use petgraph::data::FromElements;
use petgraph::graph::NodeIndex;
use petgraph::graph::UnGraph;
use petgraph::unionfind::UnionFind;
use proconio::fastout;
use proconio::input;
use proconio::marker::Chars;
use proconio::marker::Usize1;
use std::cmp::max;
use std::cmp::min;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use superslice::Ext as _;

#[fastout]
fn main() {
    input! {
        N: usize,
        mut As: [i64; N],
    }

    As.sort();

    // 1個だけマイナスにしたい場合 As[N-1] - (((As[0] - b) - c) -d)
    // 2個だけマイナスにしたい場合 (As[N-1] - ((As[0] - b) - c)) - As[1]
    // 以下同様

    // 何個マイナスにしたいか?
    let minus_cnt: usize = min(N - 1, max(1, As.iter().filter(|&a| *a < 0).count()));
    let mut outputs: Vec<String> = Vec::new();

    let mut generated = As[0] - As[minus_cnt];
    let msg0 = format!("{} {}", As[0], As[minus_cnt]).to_string();
    outputs.push(msg0);

    for i in minus_cnt + 1..N - 1 {
        let msg = format!("{} {}", generated, As[i]).to_string();
        outputs.push(msg);
        generated -= As[i];
    }

    let msg = format!("{} {}", As[N - 1], generated).to_string();
    outputs.push(msg);
    generated = As[N - 1] - generated;

    for i in 1..minus_cnt {
        let msg = format!("{} {}", generated, As[i]).to_string();
        outputs.push(msg);
        generated -= As[i];
    }

    println!("{}", generated);
    for s in outputs.iter() {
        println!("{}", s);
    }
}