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)

// Try diverta 2019 Programming Contest 2
// author: Leonardone @ NEETSDKASU

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(
        &mut std::io::stdin(),
        &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
        ($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n = get!(usize);
    let mut vs = vec![];
    for _ in 0 .. n {
        let v = get!();
        vs.push(v);
    }
    let mut op = vec![];
    vs.sort();
    
    // 最小要素 - 最大以外全プラス要素
    // 最大要素 - 最小以外全マイナス要素 
    // 最大要素 - 最小要素
    
    for i in 1 .. n-1 {
        if vs[i] < 0 {
            continue;
        }
        op.push((vs[0], vs[i]));
        vs[0] -= vs[i];
    }
    for i in 1 .. n-1 {
        if vs[i] >= 0 {
            break;
        }
        op.push((vs[n-1], vs[i]));
        vs[n-1] -= vs[i];
    }
    op.push((vs[n-1], vs[0]));
    vs[n-1] -= vs[0];
    
    println!("{}", vs[n-1]);
    for &(u, v) in op.iter() {
        println!("{} {}", u, v);
    }
}