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)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

fn solve() {
    input! {
        n: usize,
        xyv: [(i64, i64); n]
    };
    let mut xv = xyv.clone();
    let mut yv = xyv.clone();
    xv.sort_by_key(|&(x, _)| x);
    yv.sort_by_key(|&(_, y)| y);

    let mut res = 0;
    for &(x, y) in &xyv {
        let dist1 = (xv[0].0 - x).abs() + (xv[0].1 - y).abs();
        let dist2 = (xv[1].0 - x).abs() + (xv[1].1 - y).abs();
        let dist3 = (xv[n - 1].0 - x).abs() + (xv[n - 1].1 - y).abs();
        let dist4 = (xv[n - 2].0 - x).abs() + (xv[n - 2].1 - y).abs();
        let dist5 = (yv[0].0 - x).abs() + (yv[0].1 - y).abs();
        let dist6 = (yv[1].0 - x).abs() + (yv[1].1 - y).abs();
        let dist7 = (yv[n - 1].0 - x).abs() + (yv[n - 1].1 - y).abs();
        let dist8 = (yv[n - 2].0 - x).abs() + (yv[n - 2].1 - y).abs();
        res = res
            .max(dist1)
            .max(dist2)
            .max(dist3)
            .max(dist4)
            .max(dist5)
            .max(dist6)
            .max(dist7)
            .max(dist8);
    }
    println!("{}", res);
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}