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 std::collections::BTreeMap;
use std::io;

fn main() {
    let stdin = io::stdin();
    let mut buf = String::new();
    stdin.read_line(&mut buf).unwrap();
    let n: i64 = buf.trim_end().parse().unwrap();
    let mut index = BTreeMap::new();
    for _ in 0..n {
        buf.clear();
        stdin.read_line(&mut buf).unwrap();
        let mut split = buf.split_ascii_whitespace().map(|x| x.parse().unwrap());
        let x: i64 = split.next().unwrap();
        let y: i64 = split.next().unwrap();
        let (ref mut min, ref mut max) = index.entry(x).or_insert((y, y));
        if y < *min {
            *min = y;
        }
        if y > *max {
            *max = y;
        }
    }
    let first = index.iter().next().unwrap();
    let last = index.iter().next_back().unwrap();
    let ans = index
        .iter()
        .flat_map(move |(&x2, &(min, max))| {
            vec![min, max].into_iter().flat_map(move |y2| {
                vec![first, last]
                    .into_iter()
                    .flat_map(|x| vec![(x.0, (x.1).0), (x.0, (x.1).1)])
                    .map(move |(x1, y1)| (x1 - x2).abs() + (y1 - y2).abs())
            })
        })
        .max()
        .unwrap();
    println!("{}", ans);
}