Problem:
Score : 500 points

Problem Statement

There are N points on the 2D plane, i-th of which is located on (x_i, y_i).
There can be multiple points that share the same coordinate.
What is the maximum possible Manhattan distance between two distinct points?

Here, the Manhattan distance between two points (x_i, y_i) and (x_j, y_j) is defined by |x_i-x_j| + |y_i-y_j|.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq x_i,y_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the answer.

Sample Input 1

3
1 1
2 4
3 2

Sample Output 1

4

The Manhattan distance between the first point and the second point is |1-2|+|1-4|=4, which is maximum possible.

Sample Input 2

2
1 1
1 1

Sample Output 2

0

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        xy: [(i128, i128); n]
    };

    let mut xy: Vec<(i128, i128)> = xy;
    let mut v: Vec<(i128, i128, i128)> = vec![];

    for i in xy.iter() {
        v.push((i.0, i.1, i.0 + i.1))
    }
    v.sort_by(|a, b| a.2.cmp(&b.2).then(a.0.cmp(&b.0)).then(a.1.cmp(&b.1)));
    v.dedup();

    let mut min = vec![];
    let mut max = vec![];
    min.push(v[0]);
    max.push(v[v.len() - 1]);

    v.sort_by(|a, b| a.0.cmp(&b.0).then(a.1.cmp(&b.1)));
    min.push(v[0]);
    max.push(v[v.len() - 1]);

    v.sort_by(|a, b| a.1.cmp(&b.1).then(a.0.cmp(&b.0)));
    min.push(v[0]);
    max.push(v[v.len() - 1]);

    // let vc1 = v.clone();
    // let vc2 = v.clone();
    // let v_first: Vec<(i128, i128, i128)> = vc1
    //     .into_iter()
    //     .filter(|x| x.2 == first.2)
    //     .collect::<Vec<(i128, i128, i128)>>();

    // let v_last: Vec<(i128, i128, i128)> =
    //     vc2.into_iter()
    //         .filter(|x| x.2 == last.2)
    //         .collect::<Vec<(i128, i128, i128)>>();

    // let mut result = 0;
    // for i in v_first.iter() {
    //     for j in v_last() {
    //         result = std::cmp::max(result, (i.0 - j.0).abs() + (i.1 - j.1).abs())
    //     }
    // }

    let mut result = 0;
    for i in max.iter() {
        for j in min.iter() {
            result = std::cmp::max(result, (i.0 - j.0).abs() + (i.1 - j.1).abs())
        }
    }

    println!("{}", result);
}